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THE MAGNETIC CIRCUIT 



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PbbliAhed hy the 

McGraiw-Hill Book. Compcoiy 

Sixcccssora to tke&ookDeparhnentA of tke 

McGraw Publidhing Gxnpany Hill Publishing Coropony 

RiblisherA of books for 
Electrical World Hie Engineering and Mining Journal 

Engineering Record American Machinist 

Electric Railway Journal Coal Age 

Meiallurgical and Chemical Engineering R>wer 



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THE 

MAGNETIC CIECUIT 



BY 

V. KARAPETOFF 



MoGRAW-HILL BOOK COMPANY 

239 WEST 39TH STREET, NEW YOEK 

6 BouvEBiE Stbeet, London, E.C. 

1911 



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PREFACE 



This book, together with the companion book entitled " The 
Electric Circuit," is intended to give a student in electrical 
engineering the theoretical elements necessary for the correct 
understanding of the performance of dynamo-electric machinery, 
transformers, transmission lines, etc. The book also contains 
the essential numerical relations used in the predetermination of 
the performance and in the design of electrical machinery and 
apparatus. The whole treatment is based upon a very few funda- 
mental facts and assumptions. The student must be taught to 
treat every electric machine as a particular combination of electric 
and magnetic circuits, and to base its performance upon the 
fundamental electromagnetic relations rather than upon a sepa- 
rate " theory " established for each kind of machinery, as is some- 
times done. 

The book is not intended for a beginner, but for a student 
who has had an elementary descriptive course in electrical engi- 
neering and some simple laboratory experiments. The treat- 
ment is somewhat different from that given in most other books 
dealing with magnetic phenomena. It is based directly upon 
the circuital relation, or interlinkage, between an electric current 
and the magnetic flux produced by it. This relation, and 
the law of induced electromotive force, are taken to be the 
fundamental phenomena of electro-magnetism. No use what- 
ever is made of the usual artificial concepts of unitjMk, magnetic 
charge, magnetic shell, etc. These concepts of^fcthematical 
physics, together with the law of inverse squares, embody the 
theory of action at a distance, and are both superfluous and 
misleading from the modem point of view of a continuous action 
in the medium itself. 

The ampere-ohm system of units is used throughout, in 
accordance with Professor Giorgi's ideas, as is explained in the 



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vi PREFACE 

appendices. Those familiar with Oliver Heaviside's writings 
will notice his influence upon the author, in particular with 
regard to a uniform and rational nomenclature. The author trusts 
that his colleagues will judge his treatment and nomenclature 
upon their own merits, and not condemn them simply because 
they are different from the customary treatment. 

In the first four chapters the student is introduced into the 
fundamental electromagnetic relations, and is made familiar with 
them by means of numerous illustrations taken from engineering 
practice. Chapters V to IX treat of the flux and magneto- 
motive force relations in electrical machinery, first at no load, and 
then imder load when there is an armature reaction. The remain- 
ing four chapters are devoted to the phenomena of stored magnetic 
energy, namely inductance and tractive effort. The subject is 
treated entirely from the point of view of an electrical engineer, 
and the important relations and methods are illustrated by 
practical numerical problems, of which there are several hundred 
in the text. All matter of purely historical or theoretical interest 
has been left out, as well as special topics which are of interest 
to a professional designer only. An ambitious student will find 
a more exhaustive treatment in the niunerous references given in 
the text. 

Many thanks are due to the author's friend and colleague, 
Mr. John F. H. Douglas, instructor in electrical engineering in 
Sibley College, who read the manuscript and the proofs, checked 
the answers to the problems, and made many excellent sugges- 
tions for the text. Most of the sketches are original, and are the 
work of Mr. John T. Williams of the Department of Machine 
Design of Sibley College, to whom I am greatly indebted. 

Cornell University, Ithaca, N. Y., 
September, 1911. 



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CONTENTS 



PAGE 

Pbeface V 

Suggestions to Teachers xi 

Chaptbb I. The Fundamental Relation between Flux and Mag- 
netomotive Force 1 

A simple magnetic circuit. Magnetomotive force. Magnetic flux. 
The reluctance of a magnetic path. The permeance of a magnetic 
path. Reluctivity and permeability. Magnetic intensity. Flux 
density. Reluctances and permeances in series and in parallel. 

Chapter II. The Magnetic Circuit with Iron 20 

The difference between iron and non-magnetic materials. Mag- 
netization curves. Permeability and saturation. Problems involv- 
ing the use of magnetization curves. 

Chapter III. Hysteresis and Eddy Currents in Iron 32 

The hysteresis loop. An explanation of saturation and hysteresis 
in iron. The loss of energy per cycle of magnetization. Eddy cur- 
rents in iron. The significance of iron loss in electrical machinery. 
The total core loss. Practical data on hysteresis loss. Eddy cur- 
rent loss in iron. The separation of hysteresis from eddy currents. 

Chapter IV. Induced E.M.F. in Electrical Machinery 55 

Methods of inducing e.m.f. The formulae for induced e.m.f. 
The induced e.m.f. in a transformer._ The induced e.m.f. in an 
alternator and in an induction motor. The breadth factor. The 
slot factor A;,. The winding-pitch factor ky,, Nonnsinusoidal vol- 
tages. The induced e.m.f. in a direct-current machine. The ratio 
of A.C. to D.C. voltage in a rotary converter. 

Chapter V. Exciting Ampere-turns in Electrical Machinery 80 

The exciting current in a transformer. The exciting current in a 
transformer with a saturated core. The types of magnetic circuit 
occurring in revolving machinery. The air-gap ampere-turns. The 
method of equivalent permeances for the calculation of air-gap 
ampere-turns. 

vii 



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viii CONTENTS 

PAOB 

Chapter VI. Exciting Ampebe-tubns in Electrical MAcmNERT. 

(Continued) 100 

The ampere-turns required for saturated teeth. The ampere- 
turns for the armature core and for the field frame. Magnetic 
leakage between field poles. The permeance and reluctance of irreg- 
ular paths. The law of flux refraction. 

Chapter VII. Magnetomotive Force of Distributed Windings . . . 121 
The m.m.f. of a direct-current or single-phase distributed wind- 
ing. The m.m.f. of polyphase windings. The m.m.fs. in a loaded 
induction machine. The higher harmonics of the m.m.fs. 

Chapter VIII. Armature Reactign in Synchronous Machines. . . . 139 
Armature reaction and armature reactance in a synchronous 
machine. The performance diagram of a synchronous machine 
with non-salient poles. The direct and transverse armature reac- 
tion in a synchronous machine with salient poles. The Blondel 
performance diagram of a synchronous machine with salient poles. 
The calculation of the value of the coefiicient of direct reaction. The 
calculation of the value of the coeflScient of transverse reaction. 

Chapter IX. Armature Reaction in Direct-current Machines .... 163 
The direct and transverse armature reactions. The calculation 
of the field ampere-turns in a direct-current machine under load. 
Commutating poles and compensating windings. Armature reac- 
tion in a rotary converter. 

Chapter X. Electromagnetic Energy and Inductance 177 

The energy stored in an electromagnetic field. Electromagnetic 
energy expressed through the linkages of current and flux. Induc- 
tance as the coeflScient of stored energy, or the electrical inertia of a 
circuit. 

Chapter XI. The Inductance op Cables and op Transmission 

Lines 189 

The inductance of a single-phase concentric cable. The mag- 
netic field created by a loop of two parallel wires. The inductance 
of a single-phase Une. The inductance of a three-phase line with 
S3m[imetrical and semi-symmetrical spacing. The equivalent reac- 
tance and resistance of a three-phase line with an unequal spacing of 
the wires. 

Chapter XII. The Inductance op the Windings op Electrical 

Machinery 208 

The inductance of transformer windings. The equivalent leak- 
age permeance of armature windings. The leakage reactance in 
induction machines. The leakage reactance in synchronous machines. 
The reactance voltage of coils undergoing commutation 



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CONTENTS ix 

PAOB 

Chafteb XIII. The Mechanical Force and Torque Dub to Elec- 
tromagnetic Energy 240 

The density of energy in a magnetic field. The longitudinal 
tension and the lateral compression in a magnetic field. The deter- 
minatioi^ of the mechanical forces by means of the principle of virtual 
displacements. The torque in generators and motors. 

Appendix 1 262 

Appendix II 266 



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SUGIGESTIONS TO TEACHERS 



(1) This book is intended to be used as a text in a course 
which comprises lectures, recitations, computing periods, and 
home work. Purely descriptive matter has been omitted or only 
suggested, in order to allow the teacher more freedom in his 
lectures and to permit him to establish his own point of view. 
Some parts of the book are more suitable for recitations, others 
as reference in the computing room, others, again, as a basis for 
discussion in lectures or for brief theses. 

(2) Different parts of the book are made as much as possible- 
independent of one another, so that the teacher can schedule 
them as it suits him best. Moreover, most of the chapters are 
written according to the concentric method, so that it is not 
necessary to finish one chapter before starting on the next. One 
can thus cover the subject in an abridged manner, omitting the 
last parts of the chapters. 

(3) The problems given at the end of nearly every article are 
an integral part of the book, and should, under no circumstances, 
be omitted. There is no royal way of obtaining a clear under- 
standing of the underlying physical principles, and of acquiring 
an assurance in their practical application, except by the solution 
of numerical examples. It is convenient to assign each student 
the complete specifications of a machine of each kind, and ask 
him to solve the various problems in the text in application to 
these machines, in proportion as the book is covered. Numer- 
ous specifications and drawings of electrical machines will be 
foimd in the standard works of E. Arnold, H. M. Hobart, 
Pichelmayer and others, mentioned in the footnotes in the 
text. A first-hand acquaintance with these classical works on 
the part of the student is very desirable, however superficial this 
acquaintance may be. 



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XU SUGGESTIONS TO TEACHERS 

. (4) The book contains comparatively few sketches; this gives 
the student an opportunity to illustrate the important relations 
by sketches of his own. Making sketches and drawings of electric 
machines to scale, with their mechanical features, should be one 
of the important features of an advanced course, even though it 
may not be popular with some analytically-inclined students. 
Mechanical drawing develops precision of judgment, and gives 
the student a knowledge of machinery and apparatus that is 
tangible and concrete. 

(5) The author has avoided giving definite numerical data, 
coefficients and standards, except in problems, where they are 
indispensable and where no general significance is ascribed to 
such data. His reasons are: (a) Niunerical coefficients obscure 
the general exposition. (6) Sufficient numerical coefficients and 
design data will be found in good electrical hand-books and pocket- 
books, one of which ought to be used in conjunction with this 
text, (c) The student is liable to ascribe too much authority to 
a numerical value given in a text-book, while in reality many 
coefficients vary within wide limits, according to the conditions 
of a practical problem and with the progress of the art. (d) 
Most numerical coefficients are obtained in practice by assiuning 
that the phenomenon in question occurs according to a definite law, 
and by substitutuig the available experimental data into the corre- 
sponding formula. This point of view is emphasized throughout 
the book, and gives the student the comforting feeling that he 
will be able to obtain the necessary numerical constants when 
confronted by a definite practical situation. 

(6) The treatment of the magnetic circuit is made as much 
as possible analogous to that of the electrodyamic and electro- 
static circuits treated in the companion book. The teacher will 
find it advisable to make his students perfectly fluent in the use 
of Ohm's law for ordinary electric circuits before starting on the 
magnetic circuit. The student should solve several numerical 
examples involving voltages and voltage gradients, cmrents and 
current densities, resistances, resistivities, conductances, and 
conductivities. He will then find very little difficulty in master- 
ing the electrostatic circuit, and with these two the transition 
to the magnetic circuit is very simple indeed. The following 
table shows the analogous quantities in the three kinds of cir- 
cuits. 



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SUGGESTIONS TO TEACHERS 



xiu 



Electrodynamic, 

Voltage or e.m.f. 
Voltage gradient (or 
electric intensity) 



Electrostatic, 

Voltage or e.m.f. 
Voltage gradient (or 
electric intensity) 



Magnetic. 

Magnetomotive force 
M.m.f. gradient (or 
magnetic intensity) 



/ Electric current 
\ Current density 



Dielectric flux 
Dielectric flux density 



Magnetic flux 
Magnetic flux density 



r Kesistor 
i Resistance 
I Resistivity 



Elastor 

Elastance 

Elastivity 



Reluctor 

Reluctance 

Reluctivity 



{Conductor 
Conductance 
Conductivity 



Permittor (condenser) 
Permittance (capacity) 
Permittivity (dielec- 
tric constant) 



Penneator 
Permeance 
Permeability 



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LIST OF PRINCIPAL SYMBOLS 



The following list comprises most of the symbols used in the text. Those 
not occurring here are explained where they appear. When, also, a s3rmbol 
has a use different from that stated below, the correct meaning is given 
where the symbol occurs. 

Symbol. Meaning. ^Tr fi^ru^^.""^ 

a Air-gap 90 

a Width of commutator segment 237 

A Area 11 

Aa Area of flux per tooth pitch in the air 101 

At Area of flux per tooth pitch in the iron 101 

(AC) Number of ampere-conductors per centimeter 165 

b Thickness of transformer coil 211 

b Width of brush 236 

6' Thickness of mica 236 

B Flux density 14 

Bm Maximum value of the flux density 81 

C2 Number of secondary conductors 133 

Cpp Conductors per pole per phase 219 

d Duct width 94 

e, E Electromotive force 39, 65 

/ Frequency 48 

F Mechanical force 274 

Ft Tension per square centimeter 243 

Fc Compression per square centimeter 244 

H Magnetic intensity 13 

Hm Maximum value of the magnetic intensity 81 

ij I Electric cmrent 39, 205 

z'o Magnetizing current 81 

/i Current per armature branch 233 

k, k' Transformer constant 215, 221 

ka Air-gap factor 90 

ki, Breadth factor 65 

ks Slot factor 68 

kfjD Winding-pitch factor 68 

I Length 11 

XV 



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XVI LIST OF PRINCIPAL SYMBOLS 

Symbol. Meaning. Pas® ^*^®r® defined 

or first used. 

lay Ig Gross annature length 90, 102 

li Semi-net armature length 220 

In Net armature length 102 

L Inductance 184 

m Number of phases 69 

M Magnetomotive force 7 

Ma M.m.f. of armature 144 

Md M.m.f. of direct reaction 153 

Mf M.m.f. of field 144 

Mn Net m.m.f 144 

Mt M.m.f. of transverse reaction 153 

Ml Demagnetizing m.m.f 165 

Afa Distorting m.m.f 166 

n Number of turns 127, 211 

ni, Ni Number of turns in primary of a transformer 62, 81 

Ni Number of turns in secondary of a transformer 62 

N Total turns in series 65 

Om Mean length of turn 211 

p Number of poles 133 

P Power 48 

(P Permeance 9 

(Pa Permeance of air-gap 89 

(Pa Equivalent permeance around conductors in the ducts per cm. . 220 

(Pc Permeance of the path of the complete linkages 182 

(Pe' Equivalent permeance around the end-connections per cm 220 

(Peq Equivalent permeance 184 

(Pi* Equivalent permeance around embedded conductors per cm . . . 220 

(Pp Permeance of the path of the partial linkages 182 

(Pa Permeance of simplified air-gap 90 

(Pz 7Ag-z2^ permeance 223 

q Number of sections in a transformer 213 

q Number of turns per commutator segment 235 

r, R Resistance 51, 177 

(R.P.M.) Speed in revolutions per minute 260 

(R Reluctance 7 

s Distance 244 

8 Number of coils short-circuited by a brush 236 

8 Slot width 93 

S Number of slots per pole per phase 68 

S Surface area 244 

t Thickness of laminations 51 

t Time 9 

t Tooth width 93 

t' Tooth width corrected 93 

T Time of one cycle \ 64 

T Torque 253 



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UST OF PRINCIPAL SYMBOLS xvii 

Symbol. Meaning. ^^^tT^'^ 

V Velocity 59 

V Volts per ampere turn 155 

V Volume. . 39 

Wy wp Pole width 90, 166 

W Energy 39 

W Density of energy 240 

Wm Mechanical work done 250 

Wg Energy stored in the magnetic field 250 

X Reactance 144 

a Angle 69 

a Coefficient 220 

/? Phase angle 150 

y Angle 75 

d Brush shift in cm 163 

e Eddy-current constant 51 

^ Winding pitch 71 

Tj Hysteresis coefficient 48 

6 Angle 253 

X Tooth pitch 93 

fi Permeability 11 

V Reluctivity 11 

T Pole pitch 64 

<f) External phase angle ^ 144 

<b' Internal phase angle 144 

^ Flux 7 

^m Maximum value of the flux 81 

^p Flux in the path of the partial linkages 182 

^i Primary leakage flux 86 

02 Secondary leakage flux 86 

X Form factor 65 

Xa Amplitude factor 84 

(/f Angle between current and no-load e.m.f 144 

(O Angle 197 



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THE MAGNETIC CIRCUIT 



CHAPTER I 



THE FUNDAMENTAL RELATION BETWEEN FLUX 
AND MAGNETOMOTIVE FORCE 

1. A Simple Magnetic Circuit. The only known cause of 
magnetic phenomena is an electric current, or, more generally, 
electricity in motion.^ The fundamental relation between an 
electric current and magnetism can be best studied with the simple 
arrangement shown in Fig. 1. A coil, CC, of very thin wire is uni- 
formly woimd in one layer on a spool made in the shape of a circu- 
lar ring (toroid) . The tubular space inside of the ring is filled with 
some " non-magnetic '' material, so called; for instance, air, wood, 
etc. When a direct current is sent through the coil, the space 
inside the coil is found to be in a peculiar state, called the magnetic 
state. This magnetic state can be experimentally proved by 
various means, such as a compass needle', iron filings, etc. A 
region in which a magnetic state is manifested is called a magnetic 
field. Thus, in Fig. 1, the tubular space inside the coil is the mag- 
netic field excited by the current in the coil CC. 

No magnetic field is foimd in the space outside the coil upon 
exploring it with a magnetic needle or with iron filings. For 
reasons of symmetry, the field inside the ring is the same at all 
the cross-sections. Thus, a imiformly wound ring constitutes 

* Werner v. Siemens, Wiedemann's AnnaJLeUy Vol. 24. (188.5), p. 94; Lar- 
mor, Ether and Matter (1904), p. 108. The magnetism of a permanent magnet 
is probably due to molecular currents produced by some orbital motion of 
electrons within the atoms of iron. The older concepts of magnetic charges 
and free poles are summarily dismissed in this book as inadequate and 
artificial. 



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THE MAGNETIC CIRCUIT 



[Art. 1 



the simplest magnetic circuit, because the field is imiform and is 
entirely confined within the winding. 

Iron filings orient themselves within the coil in directions indi- 
cated in Fig. 1 by the concentric lines with arrow-heads. These 
lines show that the medium is " magnetized " along circles con- 
centric with the ring. Lines which show the directions in which a 
medium is magnetized are generally called magnetic lines of force} 
They are analogous to the lines of electrostatic displacement, 
though their directions and physical nature are entirely different; 



-J"^^^^^ 




SECTION A-A 

Fig. 1. — A simple magnetic circuit. 

see the chapter on the electrostatic circuit, in the author's 
Electric Circuit. 

The positive direction of the lines of force is purely conven- 
tional, and is defined as that in which the north-seeking end of a 
compass moves. Its relation to the current is, by experiment, that 
given by the right-hand screw rule. ' Namely, if the direction of 
the flow of a current is that of the rotation of a right-hand screw, 
the lines of force point in the direction of the progressive move- 
ment of the screw. Reversing the current reverses the direction 

* For actual photographs, showing iron filings which map out the magnetic 
field inside of coils of various shapes, see Dr. Benischke, Die WissenschafU 
lichen GruTidlagen der Elektrotechnik (1907), p. 126; also his Transformatoren 
(1909), pp. 4, 6, and 57. 



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Chap. I] FLUX AND MAGNETOMOTIVE FORCE 3 

of the field; this fact can be demonstrated by a small compass 
needle. The positive direction of the lines of force is indicated 
in Fig. 1 by arrow heads. The direction of the current is shown 
in the conventional way by dots and crosses; namely, a dot 
indicates that the current is approaching the observer, while a 
cross indicates that the current is receding. 

The magnetic state within the coil can also be explored by a 
small test-coU inserted into the field and connected to a galvanom- 
eter. When this coil is properly placed with respect to the field 
and then turned about its axis by some angle^ the galvanometer 
shows a deflection, because a current is induced in the coil by the 
magnetic field.* There are also other means for detecting a mag- 
netic field, for which the reader is referred to books on physics. 

The total magnetic field produced by a current is called a 
magnetic circuity by analogy with the electric and the electrostatic 
circuits. Experiment shows that the magnetic lines of force are 
always closed curves like the stream lines of an electric current, or 
like the lines of electrostatic displacement (when these are com- 
pleted through the conductors). 

Fig. 1 exemplifies a fundamental law of electromajnetism; 
namely, an electric ciu*rent creates a magnetic field in such direc- 
tions that the lines of force are linked with the lines of flow of the 
current, in the same manner that the consecutive links of a chain 
are linked together. This law admits of no theoretical proof, and 
must be accepted as a fundamental experimental fact. Wherever 
there is an electric circuit there is also a magnetic field linking 
with it. The two are inseparable, and increase and decrease 
together. Each form of an electric circuit with a certain strength 
of current in it corresponds to a definite form of magnetic field. 
It is possible that the electric current and the magnetic field 
are but two different ways of looking upon one and the same 
phenomenon. 

The linkages of magnetic lines with a current are seen more 
clearly in Fig. 11, which shows the magnetic field produced by a 
loop of wire, aa. It will be seen that the arrangement in Fig. 1 
is more suitable for an elementar}' study, because the field is much 
more uniform^ especially if the radial thickness of the ring is 
small as compared to its mean diameter, so that all the lines of 
force are of practically the same length. 

The same right-hand screw rule applies in the case of Fig. 11 



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4 THE MAGNETIC CIRCUIT [Abt. 2 

as in Fig. 1. When the current in the loop of wire circulates in 
the direction of rotation of a right-hand screw (toward the reader 
on the left), the lines of force mthin the loop point in the direc- 
tion of the progressive movement of the screw (upward). The 
rule can be reversed by saying that when the direction of the 
lines of force around a wire is that of the rotation of a right-hand 
screw, the current in the wire flows in the direction of the pro- 
gressive movement of the screw. The first statement is con- 
venient in the case of a ring winding, the second in the case of a 
long straight conductor. Both rules can be combined into one 
by considering the exciting electric circuit and the resulting mag- 
netic circuit as two consecutive links of a chain. When the arrow- 
head in one of the links (no matter which) points in the direction 
of rotation of a right-hand screw, the arrow-head in the other link, 
as it passes through the first, must point in the direction of the 
progressive movement of the screw. 

2. Magnetomotive Force. Experiment shows that the mag- 
netic field within the ring (Fig. 1) does not change if the current 
and the number of turns of the " exciting " winding vary so that 
their product remains the same. That is to say, 500 turns of 
wire with a current of 2 amperes flowing through each will pro- 
duce the same field as 1000 turns with 1 ampere, or 200 turns with 

5 amperes, because the product is equal to 1000 ampere-turns in 
all cases. Even one turn with 1000 amperes flowing through it 
will produce the same effect, provided that the turn is made of 
a wide sheet of metal spread over the whole surface of the ring, 
so as to make its action uniform throughout. 

The reason for the above can be seen by considering 1000 
separate turns with a current of 1 ampere flowing through each 
turn, and each tmn supplied with current from an independent 
electrical soiu-ce, say a dry cell. Connecting all the cells and all 
the turns in series gives 1000 turns with one ampere flowing 
through each. Connecting the cells and the turns in parallel 
results in one wide turn with 1000 amperes of ciu-rent in it. 
Such changes in the electrical connections cannot affect the action 
of each current outside the wire, because the value of the ciurent 
and the position of the turn is the same in both cases. Hence, 
the magnetic action depends only upon the number of tirnis each 
carrying 1 ampere, in other words, it depends upon the num- 
ber of ampere-turas. 



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Chap. I] FLUX AND MAGNETOMOTIVE FORCE 6 

The number of ampere-turns of the exciting winding is called 
the magnetomotive force of the magnetic circuit, because these 
ampere-turns are the cause of the magnetic field. One ampere- 
turn is the logical unit of magnetomotive force. In the example 
above, the magnetomotive force is equal to 1000 ampere-turns. 
In electric machines the field excitation often reaches several 
thousand ampere-turns, and the magnetomotive force is for con- 
venience sometimes measiu-ed in kiloampere-tums, one kilo- 
ampere-tum being equal to 1000 ampere-tmns. 

3. Magnetic Flux. The magnetic disturbance at each point 
within the ring has not only a direction, but also a magnitude. The 
disturbance is said to be in the form of a fitix, for the following 
reason: One may think of the magnetic state as being due to the 
actual displacement of some h)rpothetical incompressible sub- 
stance along the lines of force; in this case the flux represents the 
amount of this substance displaced through each cross-section of 
the ring, and is analogous to total electrostatic displacement. Or, 
as some modem writers think, there is an actual flow of an incom- 
pressible ether along the lines of force. In that case the flux may 
be thought of as the rate of flow of the ether through a cross-sec- 
tion. The viewpoint common to these two explanations gave 
rise to the name flux which means flow. 

Some physicists consider the magnetic circuit as consisting 
of infinitely subdivided (though closed) whirls or vortices in the 
ether, the rotation being in planes perpendicular to the lines of 
force. Each line of force is considered, then, as the geometric 
axis of an infinitely thin fiber or tube of force, and the ether within 
each tube in a state of transverse vortex motion. The line of 
force represents the direction of the axis of rotation, and the flux 
may be thought of as the momentum of the rotating substance 
per unit length of the tubes of force. According to any of these 
three views, the energy of a current is actually contained in the 
magnetic circuit linked with the current. 

Whichever view is adopted, the magnetic flux can be defined as 
the sum total of magnetic disturbance through a cross-section per- 
pendicular to the lines of force. Experiment shows that the total 
flux is the same through all complete cross-sections of a magnetic 
circuit. This could have been expected from the point of view 
of a displacement or flow along the lines of force; each tube of 
force being like a channel within which the displacement or the 



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6 THE MAGNETIC CIRCUIT [Abt. 3 

flow of an incompressible substance takes place. For this reason' 
the magnetic flux is said to be solenoidal (i.e., channel-shaped). 

The familiar law of electromagnetic induction discovered by 
Faraday is used for the definition of the unit of flux. Namely, when 
the total magnetic disturbance or flux within a turn of wire 
changes, an electromotive force is induced in the turn. By experi- 
ments in a imiform field, the fact is established that the value of 
the induced electromotive force is exactly proportional to the rate 
of change of the flux linking with the test loop. This fact is 
used in the definition of the unit of flux. 

With the volt and the second as the units of e.m.f. and of 
time respectively, the corresponding unit of flux is called the 
weber, and is defined as follows: A flux through a turn of wire 
changes at a uniform rate of one weber per second when the e.m.f, 
induced in the turn remains constant and equal to one volt. Such 
a unit flux can be also properly called- the voltrsecond, though as 
yet neither name has been recognized by the International Electro- 
technical Commission. The weber or the Volt-second is too large 
a imit for most practical purposes. Therefore a much smaller 
imit, called the wxixwdl} is used, which is equal to one one-hun- 
dred-millionth part of the weber, or 

one maxwell = one weber X 10" ®. 

The lines of force in Figs. 1 and 11 can be made to represent 
not only the direction of the field, but its magnitude as well, if they 
be drawn at suitable distances from each other. That is, such 
that the total number of lines passing through any part of a 
cross-section of the ring is equal numerically to the number of 
maxwells in the flux through the same part. With this conven- 
tion, each line stands symbolically for one maxwell; some engi- 
neers and physicists speak of the number of lines of force in a flux 
when they mean maxwells. 

While the weber is too large a unit, the maxwell is too small for 
many practical purposes. Therefore two other intermediate units 

* The origin of the maxwell becomes clear when one remembers that the 
volt was originally established as 10* electromagnetic C.G.S. unit of electro- 
motive force. The maxwell is related to the C.G.S. unit of e.m.f. or the so- 
called abvolt in the same way in which the weber is related to the ordinary 
volt. In other words, when the flux within a coil varies at the rate of cne 
maxwell per second, one abvolt is induced in each turn of the winding. 



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Chap. I] FLUX AND MAGNETOMOTIVE FORCE 7 

are used, namely the kilo-maxwell, equal to one thousand max- 
wells, and the mega-maxwell, equal to one million maxwells. 
These two units are sometimes called the kilo-line and the mega- 
line, the word " line " being used for the word maxwell, as 
explained above.^ 

Prob. 1. The flux within the coil (Fig. 1) is equal to 63 kilo-maxwells. 
A test coil of five turns is wound on the exciting coil so as to be linked 
with the total flux. What voltage is induced in this test coil when the 
current in the main (exciting) coil is reduced to zero at a uniform rate 
in seven seconds? Ans. 0.45 millivolt. 

Prob. 2. At what rate must the flux be varied in the preceding 
problem in order to induce one volt in the test coil? 

Ans. 0.2 weber (20 megalines) per second. 

Prob. 3. When the flux varies at a non-uniform rate show that the 
voltage induced in the test coil at any instant is equal to (d0 /dt) X 10-,* 
where t is time in seconds, and is the flux in maxwells. Show that 
the exponent of 10 must be »-2 instead of —8 if the flux is expressed, in 
megalines. 

4. The Reluctance of a Magnetic Path. Experiment shows that 
the total flux within the coil (Fig. 1) is proportional to the applied 
magnetomotive force, when the space inside is filled with air. 
Therefore, a relation similar to Ohm's law holds, namely, 

M=(R0, (1) 

where M is the magnetomotive force in ampere-turns, is the flux 
in maxwells, and (R is the coeflScient of proportionality between 
the two, called the rductance of the magnetic circuit. Script (R is 
used to distinguish reluctance from electric resistance. The mag- 
netomotive force M is the cause of the flux; or, with reference to 
an electric circuit, M is analogous to the applied electromotive 
force. is analogous to the resulting current, and the reluctance 
(R takes place of the electric resistance. Therefore, eq. (1) is 
known as Ohm's law for the magnetic circuit. Of course, the 

^This possibility of creating new units of convenient size is a great 
advantage of the metric or decimal system of units. New units are gener- 
ally understood, by the use of liatin and Greek prefixes, signifying their 
numerical relation to the fimdamental unit. For instance, it is perfectly 
legitimate to use such units as deci-ampere and hecto-volt, in spite of the 
fact that they are not in general use. Anyone familiar with the agreed 
prefixes will know that the units spoken of are equal to one-tenth of one 
ampere, and to one hundred volts. See Appendix I on the Ampere-Ohm 
System. 



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8 THE MAGNETIC CIRCUIT [Abt. 4 

analogy is purely formal, the two sets of phenomena being entirely 
different. An equation similar to (1) can be written for the flow 
of heat, of water, etc. It merely expresses the experimental fact 
that, for a certain class of phenomena, the effect is proportional 
to the cause. 

If the space within the coil be filled with practically any known 
substance, solid, liquid, or gaseous, the reluctance (R remains 
within less than ± 1 per cent of the value which obtains with air. 
The notable exceptions are iron, cobalt, nickel, manganese, chro- 
mium, and some of their oxides and alloys.^ When the circuit 
includes one of these so-called " ferro-magnetic " substances, a 
much larger flux is produced with the same m.m.f., that is, the 
reluctance of the circuit is apparently reduced to a considerable 
extent. Moreover, this reluctance is no longer constant, but 
depends upon the value of the flux. The behavior of iron and 
steel in a magnetic circuit is of great practical importance, and is 
treated in detail in Chapters II and III. 

The definition of the unit of reluctance follows directly from 
eq. (1). A magnetic circuit has a imit reluctance when a magneto- 
motive force of one ampere-turn produces in it a flux of one 
maxwell.2 No name has been given to this unit so far. The author 
ventures to suggest the name rd, and he uses it provisionally in this 
book. Granting that reluctance is a useful quantity in magnetic 
calculations, one must admit that it should be measured in some 
units of its own ; imless one chooses to use the cumbersome nota- 
tion " ampere-turns per maxwell." The name rel is simply the 
beginning of the word reluctance. Thus, a magnetic circuit has 
a reluctance of one rel when one ampere-turn produces one 
maxwell of flux in it. The imit rel is analogous to the ohm in the 
electric circuit, and to the daraf in the electrostatic circuit. 

Prob. 4. What is the reluctance of the magnetic circuit in Fig. 1 
if 47,600 ampere-turns produce a flux of 2.3 kilo-maxwells? 

Ans. 47,600/2300=20.7 rels. 

Prob. 5. How many ampere-turns are required to establish a flux 
of 1.7 megalines through a reluctance of 0.0054 rel? Ans. 9180. 

Prob. 6. A wooden ring is temporarily wound with 330 turns of 
wire; when a current of 25 amperes is flowing through the winding the 

*See Dr. C.P. Steinmetz, Magnetic Properties of Materials^ Electrical 
World, Vol. 55 (1910), p. 1209. 

2 See Appendix I at the end of the book. 



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Chap. I] FLUX AND MAGNETOMOTIVE FORCE 9 

flux is found to be equal to 21 kilo-maxwells. The permanent winding 
on the same ring must produce a flux of 65.1 kilolines at a current of 
9.3 amperes. How many turns will be required? Ans. 2750. 

6. The Permeance of a Magnetic Path. In calculations per- i 

taining to the electric circuit it is convenient to deal with the recip- 
rocals Qf resistances when conductors are connected in parallel. 
The reciprocal of a resistance is called a conductance and is meas- 
ured in mhos if resistance is measured in ohms. Similarly, a die- 
lectric is characterized sometimes by its elastance, at other times 
by the reciprocal of its elastance, which is called pwmittance. Ct^f^o^'-^i" 
When permittance is measured in farads, elastance is measured in 
daraf s (see the chapter on the Electrostatic Circuit in the author's 
Electric Circuit), 

Analogously, when two or more magnetic paths are in parallel 
it is convenient to use the reciprocals of the reluctances. The 
reciprocal of the reluctance of a magnetic path is called its per- 
meance; eq. (1) becomes then 

= (PM, (2) 

where 

(P==l/(R (3) 

A script (P is used for permeance in order to avoid confusing it 
with power. For the unit of permeance corresponding to the 
rel, the author proposes the name perm. A magnetic path has a 
permeance of one perm when one maxwell of flux is produced for 
each ampere-turn of magnetomotive force applied along the path. 

The imit " perm " has been in use among electrical designers 
for some time, although no name has been given to it. Notably 
Mr. H. M. Hobart has used it extensively in his writings, 
in the calculation of the inductance of windings. He speaks of 
''magnetic lines per ampere-turn per imit length" (of .the 
embedded part of a coil). This is equivalent to perms per unit 
length. 

In the ampere-ohm system the internationally accepted unit 
of permeance is the henry, ^ Therefore, if in eq. (2) M is measured 
in ampere-turns and in webers, (P is in henrys, and no new imit 
for permeance is necessary. In this case the reluctance (R in eqs. 

* Although the henry is defined as the unit of inductance, it is shown in 
Art. 58 below that permeance and inductance are physically of the same 
dimensions and hence measureable in the same units. 



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10 THE MAGNETIC CIRCUIT [Art. 6 

(1) and (3) is in henrys"^; or spelling the word henry backwards, 
as in the case of mho and daraf , the natural unit of reluctance in 
the ampere-ohm system gets the euphonious name of ymeh (to be 
pronounced eamey). 

Since, however, the -maxwell is used almost exclusively as the 
unit of flux, it seems advisable to introduce the rel and the perm 
as units directly related to it. Should engineers gradually feel 
inclined to use the weber and its submultiples as the imits of flux, 
then the henry, the ymeh, and their multiples and submultiples 
would naturally be used as the corresponding units of permeance 
and reluctance. 

We have, therefore, the two following systems of units for 
reluctance and permeance, according to whether the maxwell or 
the weber is used for the unit of flux (one ampere-turn being the 
unit of m.m.f . in both cases) : 



Unit of flux 


Unit of permeance 


Unit of reluctance 


Maxwell 
Weber 


Perm 
Henry 


Rel 
Ymeh 



One perm = 10" ^ henry ; one rel = 10 ® y mehs. 

Prob. 7. What is the permeance of the magnetic circuit in prob. 4? 

Ans. 0.0483 perm. « 4.83 X 10- " henry. 
Prob. 8. What is the permeance of the ring in prob. 6? 

Ans. 2.545 perm. = 0.02545 microhenry. 
Prob. 9. How many ampere-turns are required to maintain a flux 
of 2.7 megalines through a permeance of 750 perms? Ans. 3600. 

6. Reluctivity and Permeability. The reluctance of a magnetic 
path varies with the dimensions of the path according to the same 
law as the resistance of an electric conductor or the elastance of a 
dielectric. That is to say, the reluctance is directly proportional 
to the average length of the lines of force and is inversely propor- 
tional to the cross-section of the path. This relationship can be 
verified by measurements on rings of different dimensions (Fig. 1). 
When the diameter of the ring is increased twice, keeping the same 
cross-section, the length of the path of the flux is also increased 
twice. Experiment shows that the new ring requires twice as 
many ampere-tums as the first one for the same flux; or, only one- 
half of the flux is produced with the same number of ampere- 



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Chap. 1] FLUX AND MAGNETOMOTIVE FORCE 11 

turns. If the diameter of the ring is kept the same but the cross- 
section of the path is increased twice, the flux is doubled with the 
same magnetomotive force. These and similar experiments show 
that the reluctance and the permeance of a uniform magnetic path 
obey the same law as the resistance and the conductance of a con- 
ductor, or the elastance and the permittance of a prismatic slab 
of a dielectric. 

We can therefore put 

(R^vllA, (4) 

where I is the mean length of the path, A is its cross-section, and 
V is a physical constant. By analogy with resistivity and elastiv- 
ity, V is called the reliictivity of a magnetic medium. If (H is in rels, 
and the dimensions of the circuit are in centimeters, v* is in rels per 
centimeter cube. In other words, the reluctivity of a magnetic 
medium is the reluctance of a imit cube of this medium when the 
lines of force are parallel to one of the edges. For air and all other 
non-magnetic substances the experimental value of v is 0.8 rel per 
centimeter cube,^ or 0.313 rel per inch cube. 

The expression for permeance corresponding to eq. (4) is 

^=MA (5) 

where the coefiicient ji is called the permeability of the magnetic 
medium. It corresponds to the electric conductivity ;- and the 
dielectric permittivity k. Since the permeance of a path is the 
reciprocal of its reluctance, the permeability of a medium is the 
reciprocal of its reluctivity, or 

i"=Vv (6) 

When the perm and the centimeter are used for the units of per- 
meance and length, permeability is expressed in perms per centi- 
meter cube. For all non-magnetic materials /£=1.25 perms per 
centimeter cube (more accurately 1.257). With the henry and the 
centimeter as units /£= 1.257 X 10" ^ henries per centimeter cube. 
In the English system /i=3.19 perms per inch cube for non- 

*More accurately 0.796 rel per centimeter cube. As a rule, magnetic 
calculations are much less accurate than electrical calculations, because there 
is no ''magnetic insulator" known, so that there is always some magnetic 
leakage present, which is difficult to take into consideration. For this reason 
the value 0.8 is sufficiently accurate for most practical purposes. 



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12 THE MAGNETIC CIRCUIT [Art. 7 

magnetic materials. For magnetic materials fi is considerably 
larger than for non-magnetic, and varies with the field strength. 
In calculations either reluctivity or permeability is used, according 
to the conditions of the case and the preference of the engineer. 

The student has probably heard before that the permeability of 
air is assumed equal to unity. The discrepancy between this com- 
monly accepted value and the value 1.25 given above, is due to a 
different unit of magnetomotive force, called the gilbert, which is 
sometimes employed. The author considers the gilbert to be 
of doubtful utility, for reasons stated in Appendix II; hence no 
use is made of it in this book. 

Prob. 10. Assuming thd value of /£ = 1.257 to be given, check the 
value of /i»3.19 in the English system, and also the values of i; in the 
metric and the English systems, as given above. 

Prob. 11. In prob. 4 the reluctance of a ring was 20.7 rels. If the 
cross-section of the ring is 120 sq. mm., what is the average diameter of 
the ring? Ans. 9.9 cm. 

Prob. 12. How many ampere-turns are required to establish a flux 
of 47 kilolines in a ring of rectangular cross-section, made of non-magnetic 
material; the radial thickness of the ring is 8 cm., the axial width 11 cm. 
and the average radius 16 cm? Ans. About 43 kiloampere-tums. 

Prob. 13. How many ampere-turns would be required in the preced- 
ing problem for the same flux if the ring were made of iron, the relative 
permeability of which (with respect to air) is 500? 

Ans. 86 ampere-tums. 

7. Magnetic Intensity. In order that the student may better 
appreciate the significance of the concept of magnetic intensity, 
it is advisable to refresh in his mind the corresponding quantity 
used in the electric circuit, viz., the electric intensity. Namely, in 
problems on the electric and the electrostatic circuit it is some- 
times desirable to consider not only the total voltage, but also 
the voltage used up or balanced per imit length of the path along 
which the electricity fiows or is displaced. This quantity, the 
rate of change of voltage along the circuit, is known as the electric 
intensity, or the voltage gradient. It is denoted by F (see the 
Electric Circuit) ^ and is measured in volts per linear centimeter. 
When the voltage drop is imif orm along a conductor or a dielectric, 
F= jB/Z, where E is the voltage between the ends of the part of the 
circuit under consideration, and I is the corresponding length. 
When the voltage drop is not imiform, F is different for different 
points along the path, and for each point F=dE/dl. 



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Chap. I] FLUX AND MAGNETOMOTIVE FORCE 13 

In a similar way, the magnetomotive force of a magnetic circuit 
is used up bit by bit in the consecutive parts of the circuit. One 
can speak not only of the total magnetomotive force of a closed 
circuit, but also of the magnetomotive force acting upon a certain 
part of the circuit, and of magnetomotive force per unit length of 
the lines of force. Thus, for instance, if 1000 ampere-tums is con- 
sumed in a uniform magnetic circuit 4 cm. long, the magnetomo- 
tive force per unit length of path is 250 ampere-tums. 

The magnetomotive force per unit length of path is called the mag- 
netic intensity at a pointy or the m.m.f. gradient, and is denoted by 
H. Thus, if the circuit is uniform, the magnetic intensity at any 
point is 

H=M/l, (7) 

where M is the magnetomotive force acting upon the length I of 
the circuit. If the magnetic circuit is non-imiform, for instance, if 
the cross-section of the ring is different at different places, or if the 
permeability is different at some parts of the circuit due to the 
presence of iron, the m.m.f . gradient is different at different points, 
and at each point it is expressed by the equation 

H=dM/dl, . . -^ (8) 

where dM is the m.m.f. necessary for establishing the flux in the 
length dl of the circuit. If M is in ampere-tums, and Z is in centi- 
meters, /? is in ampere-tums per centimeter. 
Eqs. (7) and (8) can be also written in the form 

M=m, (9) 

and 



M- 



"£^'Hdl (10) 



These formulae, expressed in words, simply mean that the magneto- 
motive force acting upon a certain part of a magnetic circuit is the 
line integral of the magnetic intensity along the path, or the sum of 
the m.m.fs. used up in the elementary parts of the path. The rela- 
tion between M and H will become clearer to the student in the 
various applications that follow. 

Prob. 14. What is the magnetic intensity in prob. 12? 

Ans. About 425 ampere-turns per cm. of path. 



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14 THE MAGNETIC CIRCUIT [Abt. 8 

8. Fltix Density. It is often of importance to consider the jlux 
density y or the value of a flux per unit of cross-section perpendicular 
to the direction of the lines of force. Flux density is usually 
denoted by B, and is measured in maxwells (or its multiples) per 
square centimeter.^ When the flux is distributed uniformly over 
the cross-section of a path, the flux density 

B=0/A, (11) 

where A is the area of the cross-section of the path. If the flux is 
distributed non-uniformly, an infinitesimal flux d0 passing through 
a cross-section dA must be considered. In the limit, the flux den- 
sity at a point corresponding to dA is 

B=^d0/dA. (12) 

The areas A and dA are understood to be at all points normal 
to the direction of the field. Solving these two equations for the 
flux we find 

* = 5.A, (13) 

or 

^^B'dA, (14) 



■X' 



the integration being extended over the whole cross-section of the 
path. Expressed in words, these last two formulae mean that 
the total flux passing through a surface is equal to the sum of the 
fluxes passing through the different parts of that surface. 

Magnetic flux density is analogous to current density U, and 
to dielectric flux density D treated in the Electric Circuit. The 
student will find no difficulty in interpreting eqs. (11) to (14) 
from the point of view of the electric and electrostatic circuits. 

The relation between B and H is obtained from eq. (1) in which 
the value of (R is obtained from eq. (4) . Namely, we have 

M^0vl/A, 
or 

M/l=^{0/A)v. 

* Some writers express flux density in gausses, one gauss being equal to 
one maxwell per square centimeter. The unit kilogauss, equal to one kilo- 
maxwell per square centimeter, is also used. While the terms gauss and 
kilogauss are convenient abbreviations, no use is made of them in this book 
in order to keep the relation between a flux and the cross-section of its path 
explicitly before the student. 



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Chap. I] FLUX AND MAGNETOMOTIVE FOR E . 15 

The last expression, according to eqs. (7) and (11), can be written 
simply as 

H=Bv, (15) 

or, since v=l//i, 

B=iiH (16) 

Eqs. (15) and (16) state Ohm's law for a imit magnetic path, for 
instance, a path one centimeter long and one square centimeter in 
cross-section. H is the magnetomotive force between the oppo- 
site faces of the cube, /£ is the permeance of the cube, and B is the 
flux passing through it. The reader will remember similar equa- 
tions U=yF and D = kF for the imit electric^il conductor and the 
unit prism of a dielectric respectively. 

Instead of beginning the theory of the magnetic circuit with 
eq. (1) and developing it into eq. (16), it is possible to begin it with 
eq. (16). Namely, the known magnetic phenomena show that 
at each point in the medium there is a magnetic intensity H which 
is the cause of the magnetic state, and that the effect is measured 
by the flux density B; fi is the physical constant which shows the 
proportionality between H and B. The magnetic circuit is then 
assumed to be built up of infinitesimal tubes of flux in series and 
in parallel, and finally eq. (1) is obtained. 

Prob. 15. What is the flux density in prob. 12? 

Ans. 534 maxwells per square centimeter (534 gausses). 

Prob. 16. How many ampere-turns per pole are required to establish 
a flux density of 7 kflolines per square centimeter in the air-gap of a 
machine, the clearance being 3 mm.? Solution: According to eq. (15) 
H = 7000/ 1 .25 = 5600 ampere-turns per centimeter of length. Hence the 
required m.m.f. is 5600X0.3 = 1680 ampere-turns. 

9. Reluctances and Permeances in Series and in Parallel. In 

practice, one has to deal mostly with magnetic circuits of irregular 
form, for instance, those of electric machines (Fig. 24) in which 
the flux is established partly in air and partly in iron, each of vary- 
ing cross-section. The circuit consists in this case of several reluc- 
tances in series. One may say, for instance, that the total mag- 
netomotive force required in this machine, per magnetic circuit, 
is 8000 ampere-turns, of which 6000 are used in the air-gap, 1500 
in the field frame, and 500 in the armature. This is analogous to 
distinguishing between the total e.m.f . of an electric circuit, and the 
voltage drop in the various parts of the circuit. 



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16 THE MAGNETIC CIRCUIT [Art. 9 

In some cases two or more magnetic paths are in parallel, for 
instance, when there is magnetic leakage (see below). In most 
cases the engineer has to consider complicated magnetic circuits 
which consist partly of paths in series, partly of paths in parallel. 
Thus, in the same machine, the m.m.f . or the difference of mag- 
netic potential between the pole-tips is 6500 ampere-turns. This 
m.m.f. maintains a useful flux of say 2.5 megalines through the 
armature, and say 0.5 megaline of leakage or stray flux between 
the pole-tips. Thus the total flux in the field frame is 3 mega- 
lines. 

The fundamental law of the magnetic circuit, as expressed by 
eq. (1), is analogous to Ohm's law for the simple electric circuit. 
Therefore magnetic paths in series and in parallel are combined 
according to the same rule that electrical conductors are combined 
in series and in parallel. Namely, when two or more magnetic 
paths are in series, their reluctances are added; when two or more 
magnetic paths are in parallel their permeances are added. Or, 
for a series combination, 

(R^ = ^(R. . (17) 

and for a parallel Combination 

(Pe,-^(P (18) 

It will be remembered that similar relations hold also for impe- 
dances and admittances in the alternating current circuit, and 
for elastances and permittances in the electrostatic circuit. 

The proof of formulae (17) and (18) is similar to that usually 
given for the combination of electric resistances in series and in 
parallel. Namely, when reluctances are in series the total mag- 
netomotive force is equal to the sum of component m.m.f .s., or 

M^^^IM (17a) 

Dividing both sides of this equation by the common flux eq. 
(17) is obtained. When permeances are in parallel, the total flux 
is the smn of the component fluxes, or 

*e«=^^* (18a) 

Dividing both sides of this equation by the common M, eq. (18) is 
obtained. 



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Chap. I] FLUX AND MAGNETOMOTIVE FORCE 17 

One of the reasons for which calculations are as a rule more 
involved and less accurate in the magnetic than in the electric cir- 
cuit is that there is no magnetic insulation known, and therefore the 
paths of the flux in a great majority of cases cannot be shaped and 
confined at will. The student will appreciate, therefore, the reason 
for selecting a toroidal ring as the simplest magnetic circuit. If 
the winding is distributed uniformly there is no tendency for mag- 
netic leakage, except for a very sinall amount in and around each 
wire. With almost any other arrangement of a magnetic circuit 
there is a difference of magnetic potential, or an m.m.f . between 
various parts of the circuit, and part of the flux passes directly 
through the path of the least resistance, in parallel with the useful 
path. A familiar example of this is the magnetic leakage between 
the adjacent pole-tips of an electrical machine (Fig.29), or between 
the coils of a transformer (Fig. 50). 

The conditions in a magnetic circuit are similar to those 
in an imperfectly insulated electric circuit, when it, together with 
its sources of e.m.f., is immersed in a conducting liquid. Part of 
the current finds its path through the liquid instead of through 
the conductors; the current is different in different parts of the 
circuit, and the calculations are much more involved and less 
accurate, because the paths of the current in an unlimited medium 
can be estimated only approximately. 

In order to prevent or to minimize leakage the exciting ampere- 
turns should be distributed over the whole magnetic circuit, to 
each part in proportion to its reluctance. Then the m.m.f. is con- 
sumed where it is applied, and no free m.m.f . is left for leakage. 
Unfortunately, such an arrangement is impracticable in most 
cases, though it ought to be approached as nearly as possible (see 
Prob. 17 below). 

If there were a magnetic insulator, that is, a substance or a 
combination the permeability of which was many times lower than 
that of the air, it would be a great boon to the electrical industry. 
It would then be possible to avoid magnetic leakage by insula- 
ting magnetic circuits as perfectly as electric circuits are insulated. 
The absence of leakage would allow a reduction in the size of the 
field frames and exciting coils of direct- and alternating-current 
machines. It would also permit us to improve the voltage regula- 
tion of generators and transformers, to raise the power factor of 
induction motors, and to increase considerably their overload 



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18 THE MAGNETIC CIRCUIT [Art. 9 

capacity; it would also largely eliminate sparking in commutating 
machines. 



Prob. 17. A long iron rod; having a crossHsection of 9.3 sq.cm., is 
bent into a circular ring so that the ends almost touch each other. The 
ring is wound with 500 turns of wire, the winding being concentrated 
around the gap to minimize the leakage. When a current of 2.5 amperes 
is sent through the winding a flux of 74.9 kilo-maxweUs is established 
in the circuit. Assuming the reluctance of the iron to be negligible, 
calculate the clearance between the ends of the rod. 

Ans. Between 1.9 and 2.0 mm. 

Prob. 18. What is the length of the air-gap in the preceding problem 
if the estimated reluctance of the iron part of the circuit is 2 milli-rels? 

Ans. 1.7 mm. 

Prob. 19. A magnetic circuit consists of three parts, the reluctances 
of which are (Rj- 0.004 rel, (R,- 0.005 rel, and (Rs« 0.013 rel. The 
paths (R2 and (Rg are in parallel with each other and are in series with 
(Ri. What is the total permeance of the circuit? Ans. 131.4 perms. 

Prob. 20. In the preceding problem let 6I1 be the reluctance of the 
steel frame of an electric machine, (R2 be that of two air-gaps, and the 
armature, and (Rs the leakage reluctance between two poles. The ratio 
of the total flux in the frame to the useful flux through the armature 
is called the leakage factor of the machine. What is its value in this 
case? Ans. 1.38. 

Prob. 21. Referring to the two preceding problems let the air-gap be 
reduced so as to reduce the leakage factor to 1.2. How many ampere- 
turns will be required to produce a useful flux of 2.1 megaUnes in the 
magnetic circuit under consideration? Ans. 12,950. 

Prob. 22. An iron ring having a cross-section of 4 by 5 cms. is placed 
inside of a hollow ring. This ring has a mean diameter of 32 cm., an 
axial width of 11 cm., and a radical thickness of 8 cm. How many 
ampere-turns are required to produce a total flux of 47 kilollnes (count- 
ing that in the air as well as that in the iron), if the estimated relative 
permeabiHty of the iron is 1400? Hint: Let the average flux density in 
the air be Ba, and that in the iron be Bi. We have two simultaneous 
equations: 20Bi + (8S -20) Ba=- 47, and BjBa = 1400. Ans. 134. 

Prob. 23. What per cent of the total flux in the preceding problem 
is in the air? Ans. 0.24 per cent. 

Prob. 24. Show that in a ring, such as is shown in Fig. 1, the flux 
density, strictly speaking, is not uniform, but varies inversely as the 
distance from the center. Solution: Take an elementary tube of flux 
of a radius z. The magnetic intensity at any point within the tube is 
H=Ml2nXy and the flux density, according to eq. (16), B = fjLM/2nx. 

Prob. 25. What is the true permeance of a circular ring of rectangular 
cross-section, the outside diameter of which is D^ the inside diameter A, 
and the axial width h? Solution: The permeance of an infinitesimal 
tube of radius x is d(P =^ iihdxl2nx. The permeances of all the tubes 



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Chap. I] FLUX AND MAGNETOMOTIVE FORCE 19 

are in parallel and should be added; hence, integrating the foregoing 
expression between the limits iA and JA we get : (P = (ijh/2n)Ln(Di/ D2) . 

Prob. 26. Show that, when the radial thickness 6 of a ring is small as 
compared to its mean diameter D, the exact expression for permeance, 
obtained in the preceding problem, differs but little from the approxi- 
mate value, jjhb/TcDy used before. Solution: Using tjj^e expansion, 
iLn[{l+x)/{l-x)]^z+ix^+i^+ ... and puttmg 

DJD,=^(D+h)/{D-h) = (l+h/D)/{l-b/D); 
we get (P=-{ij.hb/7:D)[l+i(b/Dy+l{h/Dy+ . . . ]. When the ratio of 
& to Z) is small, all the terms within the brackets except the first one, 
can be neglected. 

Prob. 27. Show that the answer to prob. 11 is 2.1 per cent high on 
account of the density being assumed there as uniform throughout the 
cross-section of the ring. 



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CHAPTER II 
THE MAGNETIC CIRCUIT WITH IRON 

10. The Difference between Iron and Non-Magnetic Materials. 

Steel and iron differ in their magnetic properties from most other 
known materials in the following respects: 

(1) The permeability of steel and iron is several hmidred and 
even thousand times greater than that of non-magnetic materials. 

(2) The permeability of steel and iron is not constant, but 
decreases as the flux density increases. 

(3) Changes in the magnetization of steel and iron are 
accompanied by some sort of molecular friction (hysteresis) with 
the result that the same magnetomotive force produces a different 
flux when the exciting current is increasing than when it is de- 
creasing (Fig. 7). 

Besides iron, the four adjacent elements in the periodic system, 
viz,, cobalt, nickel, manganese, and chromium, are slightly mag- 
netic. Some alloys and oxides of these metals show considerable 
magnetic properties. Heusler succeeded in producing alloys of 
manganese, aluminmn, and copper which are strongly magnetic. 
These alloys have not been used in practice so far.i 

11. Magnetization Curves. The magnetic properties of the steel 
and iron used in the construction of electrical machinery are shown 
in Figs. 2 and 3. These curves are called magnetization curves, or 
B — H curves; sometimes also the saturation curves of iron. 
The flux density, in kilolines per square centimeter of -cross-sec- 
tion, is plotted, in these curves, against the ampere-turns per 
centimeter length of the magnetic circuit as abscissae. 

The student may conveniently think of these curves as represent- 

^ For the preparation and properties of Heusler's alloys see Guthe and 
Austin, BvUetim of Bureau of Standards, Vol. 2 (1906), No. 2, p. 297; Dr. C. 
P. Steinmetz, Electrical World, Vol. 55 (1910), p. 1209; Knowlton, Physical 
Reiriew, Vol. 32 (1911), p. 54. 

20 



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Chap. II] 



MAGNETIC CIRCUIT WITH IRON 



21 



ing the results of tests on samples of iron in ring form, as in Fig. 1.^ 
The current in the exciting coil is adjusted to a certain value, and 
the corresponding value of the flux in the iron ring is determined 
by any of the known means, for instance, by a discharge through 




50 100 150 

10 20 80 

H^AMPERE-TURNS PER CENTIMETER 



Scale "B" 
Scale "A" 



200 
40 



Fig. 2 — ^Magnetization in steel and iron — castings and forgings. 



*For an experimental study of the magnetic circuit with iron and for 
practical testing of the magnetic properties of steel and iron see Vol. 1, 
Chapters 6 and 7, of the author's ExperimeTUal Electrical Engineering, 



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22 THE MAGNETIC CIRCUIT [Akt. 11 

a Becondary coil connected to a calibrated ballistic galvanometer. 
The exciting ampere-turns divided by the average length of the 
path give the magnetic intensity H. The total fltix divided by the 
cross-section of the iron path gives the value of the flux density B, 
which is plotted as an ordinate against H for an absissa. Similar 
tests are made for other values of H and B\ the results give the 
magnetization curve of the material. In other words, a magneti- 
zation curve gives the relation between the magnetomotive force 
and the flux for a imit cube of the material. By combining imit 
cubes in series and in parallel a relationship is established 
between flux and ampere-turns for a circuit of any dimensions, 
made of the same material. 

The curves shown in Fig. 2 refer to the following materials: (a) 
Cast iron, which is used as the magnetic material in the stationary 
field frames of direct-current machines, and in the revolving-field 
spiders of low speed alternators. It is evident from the curves 
that cast iron is magnetically much inferior to steel; but it is used 
on account of its lower cost and ease of machining. (6) Cast steel, 
which is used fpr pole pieces, plungers of electromagnets, etc. 
It is used also for the field frames of such machines in which 
economy of weight or space is desired, for instance, in railway 
and crane motors, and in machines built for export, (c) Forged 
steel, which is used for the revolving fields of turbo-alternators, 
on accoimt of the considerable mechanical stresses developed in 
such high speed machines by the centrifugal force. 

The curves in Fig. 3 refer to carbon-steel laminations and to 
silicon-steel laminations. The former is used in the armatures of 
direct and alternating-current machines, the latter mainly in trans- 
formers. There is not much difference between the two kinds with 
regards to their B — H curves, but silicon steel shows a much lower 
loss of energy due to hysteresis and eddy currents (see Art. 20 
below). A material of much higher permeability is used for 
armature cores, when it is desired to use very high flux densities 
in the teeth. A magnetization curve for such steel laminations is 
shown in Fig. 28. 

For convenience and accuracy the lower part of each curve in 
Fig. 2 is plotted separately to a larger scale, " A," while the upper 
parts are plotted to a smaller scale, " B." Thus, Fig. 2 contains 
only three complete magnetization curves. The curve for silicon- 
steel laminations in Fig. 3 is also plotted to two different scales, 



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Chap. II] MAGNETIC CIRCUIT WITH IRON 23 

1000 ^000 8000 Scalene " 4000 




1000 2000 3000 Scale "C" 4000 

100 200 300 Scale "B" 400 

10 20 30 Scale "A" 40 

H=AMPERE.TURNS PER CENTIMETER ■ 



Fig. 3. — ^Magnetization in steel laminations. 



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24 THE MAGNETIC CIRCUIT [Abt. 12 

whfle three different scales are used for the carbonnsteel curve. 
The values of H at very low flux densities are unreliable because 
in reality each curve has a point of inflexion near the origin, not 
shown in Figs. 2 and 3 (see Fig. 7). 

The curves given in Figs. 2 and 3 represent the averages of many 
curves obtained from various sources. The iron used in an indi- 
vidual case may differ considerably in its magnetic quality from the 
average curve. The value of B obtainable with a given H depends 
to a large degree upon the chemical constitution of the specimen, 
impurities, heat treatment, etc. As a rule, the soft and pure grades 
of steel are magnetically better, that is to say, they give a higher 
flux density for the same magnetizing force, or, what is the same, 
they possess a higher permeability. Annealing improves the 
magnetic quality of iron, while pimching, hammering, etc., lowers 
it. Therefore, the laminations used in the construction of elec- 
trical machinery are usually annealed after being punched into 
their final shape. This annealing also reduces hysteresis loss. 

12. Permeability and Saturation. Permeability is defined in 
Chapter I as the permeance of a unit cube, or, according to eq. 
(16), as the ratio of B to H, The two definitions are, of course, 
identical. Therefore, the values of permeability for various values 
of B are easily obtained from the magnetization curves. For 
instance, for cast steel, at B= 15 kilolines per sq. cm. the magnetic 
intensity J? is 26 ampere-turns per cm., so that fi= 15000/26=577 
perms per cm. cube. This is the value of the absolute permeabiU 
ity in the ampere-ohm-^maxwell system. In most books the relative 
permeability of iron is employed, referring to that of the air as 
unity. Since in the above-mentioned system /£=1.25 for air, 
the relative permeability of cast steel at the selected flux density 
is 577/1.25=461. 

In practice, the calculations of magnetic circuits with iron are 
arranged so as to avoid the use of permeability pL altogether, using 
the B-H curves directly. In some special investigations, how- 
ever, it is convenient to use the values of permeability, and also 
an empirical equation between /a and B. For instance, see the 
Standard Handbook for Electrical Engineers; the topic is indexed 
" permeability — curves,'' and " permeability — equation." These 
/i— B curves show that there must be a point of inflection in 
the B—H curves at low densities, because the values of jj. reach 
their maximum at a certain definite density instead of being con- 



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Chap. II] MAGNETIC CIRCUIT WITH IRON 25 

stant for the lower part of the curves. Such would be the case if 
the lower parts of the 5— Jff curves were straight lines, as shown 
in Figs. 2 and 3, because then the ratio oi B to H would be con- 
stant. However, in ordinary engineering work the lower parts of 
magnetization curves are usually assumed to be straight lines, and 
the permeability constant. 

Three parts can be distinguished in a B — H or magnetization 
curve: the lower straight part, the middle part called the knee of 
the curve, and the upper part, which is nearly a straight line. As 
the magnetic intensity H increases, the corresponding flux density 
B increases more and more slowly, and the iron is said to approach 
saturation. With very high values of the magnetic intensity H, 
fiay several thousand ampere-turns per centimeter, the iron is com- 
pletely saturated and the rate of increase of flux density with H is 
the same as in air or in any other non-magnetic material. That is 
to say, the flux density B increases at a rate of 1.257 kilolines for 
each kilo-ampere-tum increase in H. Such is the slope of the upper 
curve in Fig. 3. 

In view of this phenomenon of saturation the total flux density 
in iron can be considered as consisting of two parts, one due to the 
presence of iron, the other independent of it, as if the paths of the 
Unes of force were in air. These two parts are shown separately 
in Fig. 4. The part OA, due to the iron, approaches a limiting 
value 5«, where the iron is saturated. The part OC, not due to the 
iron, increases indefinitely inj^ accordance with the straight line 
law, B=fiH, where /i= 1.257. The curve OD of total flux density 
resembles in shape that of OA, but approaches asymptotically 
a straight line KL parallel to OC. 

While it is customary to speak of the saturation in iron as being 
low, high, or medium, the author is not aware of any generally 
recognized method of expressing the degree of saturation numeri- 
cally. It seems reasonable to define per cent saturation in iron with 
respect to the flux density -B«, so that, for instance, the per cent 
saturation at the point N is equal to the ratio of PN' to B^. This 
method of defining saturation, while correct theoretically, pre- 
supposes that the ordinate 5« is known, which is not always the 
case. 

The percentage saturation of a machine is defined in Art. 58 of 
the Standardization Rules of the American Institute of Electrical 
Engineers (edition of 1910) as the percentage ratio of OQ to PiV, 



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26 



THE MAGNETIC CIRCUIT 



[Akt. 13 



QN being a tangent to the saturation curve at the point N under 
consideration. An objection to this definition is that according to it 
the per cent saturation does not approach 100 as N increases 
indefinitely; on the contrary, the per cent saturation gradually 
decreases to zero beyond a certain value of N. This is, of course, 
absurd. Moreover, the foregoing definition of the Institute refers 
explicitly to the " percentage of saturation of a machine," and it 
is not clear whether magnetization curves of the separate materials 
are included in it or not. The practical advantage of this definition 
as compared to that given above is that it is not necessary to 
know the value of £,. 




Fig. 4. — ^A magnetization curve analyzed. 

13. Problems Involving the Use of Magnetization Curves. 

The following problems have been devised to give the reader a clear 
understanding of the meaning of magnetization curves, and to 
develop fluency in their use. These problems lead up to the 
magnetic circuit of electric machines treated in Chapters V and 
VI. With almost any arrangement of a magnetic circuit there 
is some leakage or spreading of the lines of force, which is difficult 
to take into account theoretically. This leakage is neglected in 
most of the problems that follow, so that the results are only 
approximately correct. Leakage is considered more in detail in 
Art. 40 below, though practical designers are usually satisfied with 



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Chap. II] MAGNETIC CIRCUIT WITH IRON 27 

estimating it from the results of previous tests, rather than to 
calculate it theoretically. 

Prob. 1. Samples of cast steel are to be tested for their magnetic 
quality up to a density of 19 kilolines per square centimeter. They are 
to be in the form of rings, 20 cm. average diameter, and 0.75 sq.cm. 
cross-section. For how many ampere-turns should the exciting winding 
be designed, and what is the lowest permeance of the circuit, if some 
specimens are expected to have a permeabiHty 10 per cent lower than 
that according to the curve in Fig. 2? 

Ans. 10.4 kiloampere-tums; 1.37 perm. 

Prob. 2. Explain the reason for which it is not necessary to know the 
cross-section of the specimens in order to calculate the necessary ampere- 
turns in the preceding problem. 

Prob. 3. Some sihcon steel laminations are to be tested in the form of 
a rectangular bunch 20 by 2 by 1 cm., in an apparatus called a permeam- 
eter. The net cross-section of the iron is 90 per cent of that of the 
packet. It is found for a sample that 336 ampere-turns are required to 
produce a flux of 25.2 kilo-maxwells, the ampere-turns for the air-gaps and 
for the connecting yoke of the apparatus being eliminated. How does 
the quaUty of the specimen compare with the curve in Fig. 3? 

Ans. The permeability of the sample at 5 = 14 is about 5 per cent 
lower than that according to the curve. 

Prob. 4. What are the values of the absolute and the relative per- 
meability and reluctivity of the sample in the preceding problem? 

Ans. ft V 

relative 663 (numeric) 0.00151 (numeric) 

absolute 833 perms per cm. cube 0.(X)120 rels per cm. cube. 

Prob. 6. What is the maximum permeabiUty of cast iron according to 
the curve in Fig. 2? Ans. About 600 perms per cm. cube. 

Prob. 6. Mark in Figs. 2 and 3 vertical scales of absolute and relative 
permeability, so that values of permeabiUty could be read off directly by 
laying a straight edge between the origin and the desired point of the 
magnetization curve. 

Prob. 7. What is the percentage of saturation in carbon steel lamina- 
tions at a flux density of 20 kflo-maxwells per square centimeter, 
according to both definitions given in Art. 12? -Ajis. 92.5; 88.5. 

Prob. 8. An electromagnet has the dimensions (in cm.) shown in 
Fig. 5; the core is made of carbon steel laminations 4 nmi. thick, the 
lower yoke is of cast iron. The length of each air-gap is 2 mm.; each 
exciting coil has 450 turns. What is the exciting current for a useful 
flux of 2.2 megalines in the lower yoke? Neglect the magnetic leakage 
between the limbs of the electromagnet (this leakage is taken into consid- 
eration in the next problem). Solution: With laminations 4 mm. thick 
the space occupied by insulation between stampings is altogether negligi- 
ble; therefore the flux density in the steel is the same as in the air-gap 
and is equal to 17.2 kl/sq. cm.; in the cast iron the flux density is 11.5 



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28 



THE MAGNETIC CIRCUIT 



[Art. 13 



kl/sq. cm. One-half of the average length of the path in the steel is 
37.3 cm., and in the cast iron 20.5 cm. Hence, with reference to the 
magnetization curves, we find for one-half of the magnetic circuit (the 
other half being identical, it is sufficient to calculate for one-half) : 

amp.-tums for steel core 65 X 37.3 = 2425 

amp.-tums for one air-gap 0.2 X 0.8 X 1 7200 = 2752 

amp.-turns for the cast-iron yoke 180 X 20.5 =« 3690 



Total 8867. 

Ans. The exciting current is 8867/450 = 19.7 amperes. 
Prob. 9. In the solution of the preceding problem the effect of leakage 
is disregarded. It is found by experiments on similar electromagnets 



^ ift A 




y 



Fig. 5. — ^An electromagnet (dimensions in centimeters). 

that the leakage factor is equal to about 1.2, that is to say, the flux in the 
upper yoke is 20 per cent higher than that in the lower one. This means 
that out of every 1200 lines of force in the upper yoke 1000 pass through 
the lower yoke as a part of the useful flux, and 2(X) find their path as a 
leakage through the air between the limbs, as shown by the dotted lines. 
Calculate the exciting current required in the preceding problem, assuming 
(a) that the total leakage flux is concentrated between the two air-gaps 
along the line (m\ (h) that it is concentrated along the line 66, at one- 
third of the distance fnom the bottom of the exciting coil, that is 6.33 
cm. from the air-gaps. 

Ans. (a) 44.2 amperes; (6) 40 amperes. 
Prob. 10. Show that it is more correct in the preceding problem to 
assume the leakage flux concentrated at one-third of the distance from 
the bottom of the exciting coils, than at the center of the coils. 



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Chap. II] MAGNETIC CIRCUIT WITH IRON 29 

Prob. 11. A ring of forged steel has such dimensions that the average 
length of the lines of force is 70 cm. The ring has an air-gap of 1 .5 nmi., and 
is provided with an exciting winding concentrated near the air-gap so as 
to minimize the leakage. What is the flux density at an m.m.f. of 4000 
ampere-turns? First Solution : Assume various values of B, calculate the 
corresponding values of the ampere-turns, until the value of 5 is found, for 
which the required excitation is 4000 ampere-turns (solution by trials). 
Second solution: Let the unknown density be B and the corresponding 
magnetic intensity in the steel be H. The required excitation for the steel 
is then 70H, and for the air-gap 0.15X0.8X10005 = 1205 ampere-turns. 
Therefore, 

70£f + 1205 =4000. 

The values of B and H must satisfy this equation of a straight line,and 
besides they must be related to each other by the magnetization curve 
for steel forgings (Fig. 2). Hence, B and H are determined by the intersec- 
tion of the straight line and the curve. The straight line is determined by 
two of its points ; for instance, when -ff = 40, 5 « 10 ; when H = 24, 5 = 19.3. 
Drawing this line in Fig. 2 we find that the point of intersection corre- 
sponds to 5 = 16.3.* Ans. 16.3 kilolines per sq. cm. 

Prob. 12. Solve the preceding problem, assuming the ring to be made 
of silicon steel laminations: 10 per cent of the space is taken by the 
insulation between the laminations. 

Ans. . Flux density in the laminations is 15.2 kl/sq. cm. 

Prob. 13. In a complex magnetic circuit, an air^ap 3 mm. long and 
26 sq. cm. in cross-section is shunted by a cast-iron rod 14 cm. long and 
10 sq. cm. in cross-section. What is the number of ampere-turns neces- 
sary for producing a total flux of 215 kilolines through the two paths in 
parallel, and what is the reluctance of the rod per centimeter of its length 
imder these conditions? Ans. 1160 ampere-turns; 0.933 milli-rel. 

Prob. 14. The magnetic flux in a closed iron core must increase and 
decrease according to a straight-line law with the time, then reverse and 
increase and decrease according to the same law in the opposite direction. 
Show the general shape of the curve of the exciting current, neglecting 
the effect of h3rteresis. 

Prob. 16. Show that if in the preceding problem the flux varies accord- 
ing to the sine law the curve of the exciting current is a peaked wave. 
Show how to determine the shape of this curve from a given magnetiza- 
tion curve of the material. This problem has an application in the calcu- 
lation of the exciting current in a transformer. 

Prob. 16. In the magnetic circuit shown in Fig. 6 the useful flux 
passes through the air-gap between the two steel poles; a part of the flux 

* The student will see from the solution of this problem that in the case 
of a series magnetic circuit it is much easier to find the m.m.f . required for 
a given flux than vice versa. On the other hand, in the case of two mag- 
netic paths in parallel (such as in prob. 13), it is easier to find the flux for a 
given m.rni. 



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30 



THE MAGNETIC CIRCUIT 



(Art. 13 



is shunted through the cast-iron part of the circuit. At low saturations a 
considerable part of the total flux is shunted through the cast-iron part, 
but as the flux density increases the cast iron becomes saturated, and a 
larger and larger portion of the flux is deflected into the air-gap. What 
percentages of the total flux in the yoke are shunted through the cast iron 
when the flux density in the air-gap is 1 kl/sq. cm. and 7 kl/sq. cm. 
respectively? Solution : When the flux density in the air-gap is 1 kilo- 
Une per sq. cm. the m.m.f. across the gap is 1000 X 0.8 X 0.5 «» 400 ampere- 
turns. The flux density in the steel poles in 2 kl/sq. cm., and the required 
m.m.f. in them is about 16 ampere-turns. Therefore, the total m.m.f. 
across AC and consequently across the cast-iron part is 416 ampere-turns. 




36 Sq. Cm. 
Caaifr Steel 




Fig. 6. — A complex magnetic circuit. 

or H = 24.5 ampere-turns per centimeter of length of the path in the cast 
iron. This value of H corresponds on the magnetization curve to5 =6kl/sq. 
cm. ; hence, the total flux in the cast iron is 72 kl. The flux in the yoke is 
60+72 = 132 kl., and the percentage in the cast-iron shunt is 72/132 or 
about 55 per cent. Similarly, it is found that, when the flux density in the 
air-gap is 7/kl sq. cm., about 25 per cent of the flux is shunted through 
the cast-iron part. The foregoing arrangement illustrates the principle 
used in some practical cases, when it is desired to modify the relation 
between the flux and the magnetomotive force, by pro\dding a highly 
saturated magnetic path in parallel with a feebly saturated one. 

Ans. 55 per cent and 25 per cent approximately. 
Prob. 17. Indicate how the preceding problem can be solved if the 
cast-iron part were provided with a small clearance of say 1 mm. Hint : 
See the second solution to problem 11. 



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Chap. II] MAGNETIC CIRCUIT WITH IRON 31 

Prob. 18. What is the length of the yoke in Fig. 6 if the exciting cur- 
rent increases 12 times when the flux density in the air-gap increases from 
1 to 7 kl/sq. cm.? Hint: If Hi and H, are the known magnetic intensi- 
ties in the yoke, corresponding to the two given densities, and x is the 
unknown length of the yoke, we have, using the values obtained in the 
solution of problem 15 : (HiX + 416) 12 - H^ + 3090. 

Ans. About 1.2 m. 



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CHAPTER m 
HYSTERESIS AND EDDY CURRENTS IN IRON 

14. The Hysteresis Loop. Steel and iron possess a property 
of retaining part of their magnetism after the external magnetomo- 
tive force which magnetized them has been removed. Therefore, 
the magnetization or the B-H curve of a sample depends some- 
what upon the magnetic state of the specimen before the test. This 
property of iron is called hysteresis. The curves shown in Figs. 2 
and 3 refer to the so-called virgin state of the materials, which state 
is obtained by thoroughly demagnetizing the sample before the 
test. A piece of iron can be reduced to the virgin state by placing 
it withm a coil through which an alternating current is sent, and 
gradually reducing the current to zero. Instead of changing the 
current, the sample can be removed from the coil. 

Let h sample of steel or iron to be tested be made into a ring 
and provided with an exciting winding, as in Fig. 1. Let it be 
thoroughly demagnetized; in other words, let its residual mag- 
netism be removed; then let the ring be magnetized gradually or in 
steps to a certain value of the flux density. Let OA in Fig. 7 rep- 
resent the virgin magnetization curve, that is to say the relation 
between the calculated values of B and H from this test, and let 
PA be the highest flux density obtained. If now the magnetizing 
current be gradually reduced, the relation between B and H is no 
more represented by the curve OA, but by another curve, such as 
AC] this is because of the above-mentioned property of iron to 
retain part of its magnetism. When the current is reduced to 
zero, the specimen still possesses a residual flux density OC, Let 
the current now be reversed and increased in the opposite direc- 
tion, imtil H reaches the negative value OF, at which no magnetic 
flux is left in the sample. The value of H=OF is called the coer- 
cive force. When the magnetic intensity reaches the negative value 
of OP^ = OPj experiment shows that the magnetic density P'A' in 
the sample is equal and opposite to PA. 

32 



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Chap. Ill] 



HYSTERESIS AND EDDY CURRENTS 



33 



Let now the exciting current be again decreased, reversed and 
increased to its former maximum value corresponding to H=OP, 
It will be f oimd that the relation between B and H follows a differ- 
ent though symmetrical curve, A'C'F'A, which connects with the 
upper curve at the point A. The complete closed curve is called 
the hysteresis loop; a sample of iron which has been subjected to a 
varying magnetomotive force as described before, is said to have 
undergone a complete cycle of magnetization. If the same cycle 



Q 


3 

A 








11. 

a. 


Y 




A' 


J\ 


Q' 





Fig. 7. — ^A hysteresis loop. 

be repeated any number of times, the curve between B and K 
remains the same, as long as the physical properties of the sample 
remain unchanged. 

The lower half of the hysteresis loop is identical with the 
inverted upper half, so that the residual flux density OC'=OC, 
and the coercive force OF'==OF, The shape of the loop for a 
given sample is completely determined by the maximum ordinate 
AP, or the maximum excitation OP, If the excitation be carried 
further, for instance, to the point D on the virgin curve, the 
hysteresis loop would be larger, beginning at the point D, and 
would be similar in its general shape to the loop shown in Fig. 7. 



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34 THE MAGNETIC CIRCUIT [Abt. 15 

A piece of iron can also be carried through a hysteresis cycle 
mechanically. Thus, instead of changing the excitation, the 
sample may be moved to a weak field, reversed, and returned to 
its original location. The relation between B and H, however, 
will be the same in either case. 

An important feature of the hysteresis cycle is that it requires 
a certain amoimt of energy to be supplied by the magnetizing 
current, or by the mechanism which reverses the iron with 
respect to the field. It is proved in Art. 16 below that this energy 
per cubic imit of iron is proportional to the area of the hysteresis 
loop. This energy is converted into heat in the iron, and therefore 
from the point of view of the electromagnetic circuit represents 
a pure loss. If the cycles of magnetization are performed in 
sufficiently rapid succession, for instance by using alternating 
current in the exciting winding, the temperature of the iron rises 
appreciably. 

The phenomenon of hysteresis is irreversible; that is to say, 
it is impossible to make a piece of iron to undergo a cycle of mag- 
netization in the direction opposite to that indicated by arrow- 
heads, in Fig. 7. If it were reversible the loss of energy occasioned 
by performing the cycle in one direction could be regained by 
performing it in the opposite direction. In this respect the 
hysteresis cycle differs materially from the theoretical reversible 
cycles studied in thermodynamics, and reminds one of an irre- 
versible thermodynamic cycle, in which friction or sudden expan- 
sion is present. 

16. An Explanation of Saturation and Hysteresis in Iron. 
While the physical nature of magnetism is at present unknown, 
there is sufficient evidence that the magnetization of iron is 
accompanied by some kind of molecular change. Let us assume, 
in accordance with the modem electronic theory, that there is an 
electric current circulating within each molecule of iron, due to the 
orbital motion of one or more electrons within the molecule. Each 
molecule represents, therefore, a minute electromagnet acted upon 
by other molecular electromagnets. In the neutral state of a piece 
of iron, the grouping of the molecules is such that the currents 
are distributed in all possible planes, and the external magnetic 
action is zero. Under the influence of an external magnetomotive 
force the molecules are oriented in the same way that small mag- 
netic needles are deflected by an external magnetic field. With 



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Chap. Ill] HYSTERESIS AND EDDY CURRENTS 35 

small intensities of the external field, the molecules of iron return 
into their original stable positions as soon as the external m.m.f. 
is removed; when, however, the external magnetic intensity 
becomes considerable some of the molecules turn violently and 
assume new groupings of stable equilibrium. Therefore, when 
the external m.m.f. is removed, there is some intrinsic magneti- 
zation left, and we have the phenomenon of residual mag- 
netism. 

With an ever-increasing external m.m.f., more and more of the 
molecules are oriented so that their m.m.fs. are in the same direc- 
tion as the external field, the iron then approaching saturation. 
Any further increase in the flux density is then mainly due to the 
flux between the molecules, the same as in any non-magnetic 
medium. 

According to the foregoing theory, an external m.m.f. turns 
the internal m.m.fs. into more or less the same direction; these 
m.m.fs. then help to establish the flux in the intermolecular spaces 
which are much greater than the molecules themselves. There- 
fore, the higher flux density in iron is not due to a greater permea- 
bility of the iron itself, but to an increased m.m.f. It is never- 
theless permissible, for practical purposes, to speak of a higher 
permeability of the iron, disregarding the internal m.m.fs., and 
considering the permeability, according to eq. (16), as the ratio 
of the flux density to the externally applied magnetic intensity. 

The foregoing theory explains also the general character of 
the permeability curve of iron. With very small values of H 
the molecules of a piece of iron are oriented. but very little, but 
are rapidly oriented more and more as H is increased. There- 
fore, for small values of H^ jj. must be expected to increase with H, 
On the other hand, when the saturation is very high, an increase 
in H changes B but little, because practically all of the available 
internal m.m.fs. have been utilized. Therefore, for large values 
of Hj ji decreases with increasing H, Consequently, there is a 
value of H for which /i is a maximum. This is the actual shape of 
permeability curves (see for instance the reference to the Standard 
Handbook given in Art. 12 above). 

The phenomenon of magnetization is irreversible because the 
changes from one stable grouping of molecules to the next are 
sudden. Each molecule, in changing to a new grouping, acquires 
kinetic energy, and oscillates about its new position of equilib- 



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36 



THE MAGNETIC CIRCUIT 



[Akt. 15 



rium until the energy is dissipated by being converted into heat. 
This heat represents the loss of energy due to hysteresis. 

This theory of satura-tion and hysteresis is due originally 
to Weber, and has been improved by Ewing, who has shown 
experimentally the possibility of various stable groupings of 
a large number of small magnets in a magnetic field. By varying 
the applied m.m.f . he obtained a curve similar to the hysteresis 
loop of a sample of iron. For further details of this theory see 
Ewing, Magnetic Ivduction in Iron and other Metals (1892), 
Chapter XI. 

The following analogy is also use^. Let a body Q (Fig. 8), 
rest on a support and be held in its central position by two springs 




Fig. 8. — ^A mechanical analogue to hysteresis. 

S, Sy which can work both under tension and imder compression. 
Let this body be made to move periodically to the right and to 
the left of its central position, under the influence of an alterna- 
ting external force H, Call B the deflections of the body from 
its middle position. The relation between B and H is then similar 
to the hysteresis loop in Fig. 7, provided that there is some 
friction between the body Q and its support, and provided that 
the springs offer in proportion more resistance when distorted 
greatly than when distorted slightly. 

Starting with the neutral position of the body let a gradually 
increasing force H be applied which moves the body to the right. 
This corresponds to the virgin curve in Fig. 7, except that this 
simple analogy does not account for the inflection in the virgin 
curve near the origin. Let then the force H be gradually reduced, 
allowing the springs to bring Q nearer the center. When the 



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Chap. Ill] HYSTERESIS AND EDDY CURRENTS 37 

external force is entirely removed, the body is still somewhat to 
the right of its central position, because the friction balances part 
of the tension of the springs. Here we have something analogous 
to residual magnetism and to the part AC of the hysteresis loop. 
A finite force H is required in the negative direction to bring Q 
to the center. This force corresponds to the coercive force of a 
piece of iron. 

By following this analogy through the complete cycle one 
can show that a loop is obtained similar to a hysteresis loop. 
Also, it can be shown that the phenomenon is irreversible, and 
that total work done by the force H is equal to the work of friction. 
Moreover there is a periodic interchange of energy between the 
springs and the source of the force -ff , and the net loss of energy 
is represented by the area of the loop corresponding to Fig. 7. 

Prob. 1. An iron ring is thoroughly demagnetized, and then the cur- 
rent in the exciting winding is varied in the following manner: It Is 
increased gradually from zero to 1 ampere and is then reduced to zero. 
After this, the current is increased to 2 amperes in the same direction, and 
again reduced to zero. Then the current is increased to 3 amperes again in 
the same direction, and reduced to zero, etc. Draw roughly the general 
character of the B-H curve, taking the hysteresis into consideration. 
Hint: First study a similar process on the mechanical analogy shown in 
Fig. 8.» 

Prob. 2. A piece of iron is made to undergb a magnetization process 
from the point A (Fig. 7) to a point between F and A' such that, when 
subsequently the exciting circuit is opened, the ascending branch of the 
hysteresis curve comes to the origin. Show that such a process does not 
bring the iron into the neutral virgin state, in spite of the fact that J5 =0 
for i/=0. fiint: Consider the further behavior of the iron for positive 
and negative values of H, 

Prob. 3. A millivoltmeter is connected to the high-tension terminals 
of a transformer, and the current in the low-tension winding is varied in 
such a way as to keep the voltage constant: Show that the curve of the 
current plotted against time is proportional to the hysteresis loop of the 
core. Hint: Since d^/dt is constant, (P is proportional to the time. 

Prob. 4. The magnetic flux density in an iron core is to vary with the 
time according to the sine law. Plot to time as abscissse the instantane- 
ous values of the exciting ampere-turns per centimeter length of the core 
from an available hysteresis loop, and show that the wave of the exciting 
current is not a sine wave and is unsjmametrical. Note: This problem 
has an application in the calculation of the exciting current of a trans- 
former; see Art. 33 below. 

* A solution of this and of the next problem will be found in Chapter V 
of Ewing's Magnetic Induction in Iron and other Metala, 1892. 



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38 THE MAGNETIC CIRCUIT [Art. 16 

16. The Loss of Energy per Cycle of Magnetization. When a 
magnetic flux is maintained constant the only energy supplied 
from the source of electric power is that converted into the i^r 
heat in the exciting winding; no energy is necessary to maintain 
the magnetic flux. This is an experimental fact, fundamental in 
the theory of magnetic phenomena. When, however, the flux 
is made to vary, by varying the exciting ampere-turns or the 
reluctance of the magnetic circuit, electromotive forces are 
induced in the magnetizing, winding by the changing flux. 
A transfer of energy results between the electric and the magnetic 
circuits. 

Beginning, for instance, at the point A of the cycle (Fig. 7), . 
and going toward C, the flux is forced to decrease. According 
to Faraday's law, the e.m.f . induced by this flux in the magnetiz- 
ing winding is such as to resist the change, i.e., it tends to main- 
tain the current. Therefore, during the part AC of the cycle 
energy is supplied from the magnetic to the electric circuit. 
This shows that energy is stored in a magnetic field. During the 
part CFA' of the hysteresis loop energy is supplied from the 
electric to the magnetic circuit, because at the point C, the 
current is reversed and becomes opposed to the e.m.f. The 
other half of the cycle being symmetrical, with the flux and the 
current reversed, energy is returned to the electric circuit during 
the part A'C of the cycle, and is again accumulated in the 
magnetic circuit during the part C'F'A. 

If the part AC of the cycle were identical with C'F'A^ and the 
part A^C were identical with CFA\ the amounts of ^ergy trans- 
ferred both ways would be the same, and there would be no net 
loss of energy at the end of the cycle. In reality the two parts 
are different; the amoimts.of energy returned from the magnetic 
circuit to the electric circuit in the parts AC and A'C are smaller 
than the amounts supplied by the electric circuit in the parts 
CFA' and C'F'A, This is because the last two parts of the curve 
are more steep than the first two, and consequently the induced 
e.m.fs. are larger for the same values of the current. The net 
result is therefore an input of energy from the electric into the 
magnetic circuit, this enjergy being converted into heat in the iron. 
No such effect is observed with non-magnetic materials, because 
the two branches of a complete B-H cycle coincide with a straight 
line passing through the origin. 



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Chap. IIII HYSTERESIS AND EDDY CURRENTS 39 

To prove that the energy lost per cubic unit of iron per cycle 
of magnetization is represented by the area of the hysteresis loop, 
we first write down the expression for the energy returned to the 
electric circuit during an infinitesimal change of flux in the 
part AC of the cycle. Let the flux in the ring at the instant under 
consideration be (P webers, and the magnetomotive force ni amp- 
ere-turns, where i is the instantaneous value of the current, and n 
is the total number of turns on the exciting winding. The instan- 
taneous induced e.m.f ., due to a decrease of the flux by d^ during 
an infinitesimal element of time dt seconds, is e= —nd^/dt volt. 
The sign minus is necessary because e is positive (in the direction 
of the current) when d0 is negative, that is to say, when the flux 
decreases. The electric energy corresponding to this voltage is 

dW=eidt= —nid^ watt-seconds (joules). 

Hence, the total energy returned to the electric circuit during the 
part AC of the cycle is 



W 



= — r nid^, 



or, interchanging the limits of integration, 

XA 
nid0. 

Since all the parts of the ring undergo the same process, and 
the curve in Fig. 7 is plotted for a unit cube of the material, it is 
of interest to find the loss of energy per cubic centimeter of mate- 
rial. If S is the cross-section and I the mean length of the lines 
of force in the iron, we have that the volume 

V^Sl cubic centimeters. 

Dividing the expression for the energy by this equation, we find 
that the energy in watt-seconds per cubic centimeter of iron is 

W/V=^\in/l)'{d0/S)^C^HdB, . . . (19) 

where H is in ampere-turns per centimeter, and B is in webers 
per square centimeter. 

But HdB is the area of an infinitesimal strip, such as is shown 
by hatching in Fig. 7. Consequently, the right-hand side of eq. 



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40 THE MAGNETIC CIRCUIT [Art. 17 

(19) represents the area of the figure ACQ, which is therefore a 
measure for the energy transferred to the electric circuit, per cubic 
centimeter. In exactly the same way it can be shown that the 
energy supplied to the magnetic circuit during the part C'A of the 
cycle is represented by the area ACQ. Hence the net energy 
loss for the part of the cycle to the right of the axis of ordinates 
is represented by the area ACC'A. Repeating the same reasoning 
for the left-hand side of the loop it will be seen that the total 
energy loss per cycle of magnetization per cubic centimeter of 
material is represented by the area ACA'C'A of the hysteresis 
loop. For a given material, this area, and consequently the loss, 
is a fimction of the maximum flux density PA, and increases with 
it according to a rather complicated law. Two empirical formulae 
for the loss of energy as a fimction of the density are given in Art. 
20 below. 

In the problems that follow the weight of one cubic decimeter 
of solid carbon steel is taken to be 7.8 kg., and that of the alloyed 
or silicon steel 7.5 kg. The weight of one cubic decimeter of 
assembled carbon steel laminations is taken as 0.9X7.8 = 7 kg., 
and that of silicon steel laminations as 0.9X7.5=about 6.8 kg. 

Prob. 6. A hysteresis loop is plotted to the following scales: abscissae 
1 cm. = 10 amp.-tums/cm. ; ordinates, 1 cm. = 1 kilo-maxwell /sq. cm.; the 
area of the loop is found by a planimeter to be 72 sq. cm. What is the 
loss per cycle per cubic decimeter of iron? 

Ans. 7. 2 watt-seconds (joules). 

Prob. 6. The hysteresis loop mentioned in the preceding problem was 
obtained from an oscillographic record at a frequency of 60 cy., with a sam- 
ple of iron which weighed 9.2 kg. What was the power lost in hysteresis 
in the whole ring? Ans. 510 watts. 

Prob. 7. The stationary coil of a ballistic electro-dynamometer is con- 
nected in series with the exciting electric circuit (Fig. 1) ; the moving coil 
is connected through a high resistance to a secondary winding placed on 
the ring. The exciting current is brought to a certain value, and then the 
current is reversed twice in rapid succession, in order that the iron may 
undergo a complete magnetization cycle. Show that the deflection of the 
electro-dynamometer is a measiu^ for the area of the hysteresis loop. 
Hint: HdB^H(dB/dt) (ft-Const. X iedt.' 

17. Eddy Currents in Iron. Iron is an electrical conductor; 
therefore when a magnetic flux varies in it, electric currents are 

'Searie, *'The Ballistic Measurement of Hysteresis," Electrician, Vol. 
49, 1902, p. 100. 



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Chap. Ill] HYSTERESIS AND EDDY CURRENTS 41 

induced along closed paths of least resistance linked with the flux. 
These currents permeate the whole bulk of the iron and are called 
eddy or Foucault currents. Eddy currents cause a loss of energy 
which must be supplied either electrically or mechanically from 
an outside source. Therefore, the iron cores used for variable fluxes 
are usually built of laminations, so as to limit the eddy currents 
to a small amoimt by interposing in their paths the insulation 
between the laminations. Japan, varnish, tissue paper, etc., are 
used for this purpose. In many cases the layer of oxide formed 
on laminations during the process of annealing is considered to be 
a sufficient insulation against eddy currents. 

The usual thickness of lamination varies from 0.7 to 0.3 mm., 
according to the frequency for which an apparatus is designed, 
the flux density to be used, the provision for cooling, etc. The 
more a core is subdivided the lower is the loss due to eddy currents, 
but the more expensive is the core on account of the higher cost 
of rolling sheets, and of pimching and assembling the laminations. 
Besides, more space is taken by insulation with thinner stampings, 
so that the per cent net cross-section of iron is reduced. The 
net cross-section of laminations is usually from 95 to 85 per cent 
of the gross crossnsection, depending upon the thickness of the 
laminations, the kind of insulation used, and the care and pres- 
sure used in assembling the core. For preliminary calculations 
about ten per cent of the gross cross-section is assumed to be 
lost in insulation. 

Fig. 9 shows two iron cores in cross-section, one core solid, 
the other subdivided into three laminations by planes parallel 
to the direction of the lines of force. The lines of force are shown 
by dots, and the paths of the eddy currents by continuous lines. 
Eddy currents are linked with the lines of force, the same as the 
current in the exciting winding. In fact, eddy currents are similar 
to the secondary currents in a transformer, inasmuch as they 
tend to reduce the flux created by the primary current. The 
core must be laminated in planes perpendicular to the lines of 
flow of the eddy currents, so as to brea,k up their paths and at the 
same time not to interpose air-gaps in the paths of the lines of 
force. 

An iron core can be further subdivided by using thin iron 
wires in place of laminations. Such cores were used in early 
machines and transformers, but were abandoned on accoimt of 



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42 



THE MAGNETIC CIRCUIT 



[Art. 18 



expense and poor space factor. Iron-wire cores are used at present 
in only high-frequency apparatus, in which eddy currents must 
be carefully guarded against; for instance in the induction coils 
(transformers) employed in telephone circuits. 

It will be seen by an inspection of Fig. 9 that eddy currents 
are much smaller in the laminated core because the resistance of 
each lamination is increased while the flux per lamination and con- 
sequently the induced e.m.f . is considerably reduced. It is proved 
in Art. 21 below that the power lost in eddy currents per kilogram 
of laminations is proportional to the square of the thickness of 




Fig. 9. — Eddy currents in a solid and in a laminated core. 

the laminations, the square of the frequency, and the square of 
the flux density. 

Prob. 8. Show that the armature cores of revolving machinery must 
be laminated in planes perpendicular to the axis of rotation. 

Prob. 9. Show that assuming the temperature-resistance coefficient of 
iron laminations to be 0.0046 per degree Centigrade the eddy current loss of 
a core at 70° C. is only about 82.5 per cent of that at 20° C. 

Prob. 10. Explain the reason for which the hysteresis loss in a given 
core and at a given frequency depends only on the amplitude of the excit- 
ing current, while the eddy-current loss depends also upon the wave-form 
of the current. 

18. The Significance of Iron Loss in Electrical Machinery. 

The power lost in an iron core on accoimt of hysteresis and eddy 
currents, taken together, is called iron loss or core loss. It is of 
importance to imderstand the effect of this loss in the iron cores 



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Chap. Ill] HYSTERESIS AND EDDY CURRENTS 43 

of electrical machineiy and apparatus: First, because they bring 
about a loss of power and hence lower the efficiency of a 
machine; Secondly, because they heat up the iron and thus 
limit the permissible flux density, or make extra provisions for 
ventilation and cooling necessary; Thirdly, because they afifect 
the indications of measuring instruments. The effects of hystere- 
sis and eddy currents in the principal types of electrical machinery 
are as follows: 

(a) In a transformer an alternating magnetization of the iron 
causes a core loss in it. The power thus lost must be supplied from 
the generating station in the form of an additional energy com- 
ponent of the primary current. The core is heated by hysteresis 
and by eddy currents, and the heat must be dissipated by the 
oil in which the transformer is immersed, or by an air blast. 

(6) In a direct-current machine the revolving armature is 
subjected to a magnetization first in one direction and then in 
the other; the heating effect due to the hysteresis and eddy 
currents is particularly noticeable in the armature teeth in which 
the flux density is usually quite high. The core loss, being sup- 
plied mechanically, causes an additional resisting torque between 
the armature and the field. In a generator this torque is sup- 
plied by the prime mover; in a motor this torque reduces the 
available torque on the shaft. 

(c) The effect of hysteresis and of eddy currents in the armature 
of an alternator or of a synchronous motor is similar to that in a 
direct-current machine. 

(d) In an induction motor the core loss takes place chiefly 
in the stator iron and teeth, where the frequency of the magnetic 
cycles is equal to that of the power supply ; the frequency in the 
rotor corresponds to the per cent slip, so that even with very 
high flux densities in the rotor teeth the core loss in the rotor is 
comparatively small. At speeds below synchronism the necessary 
power for supplying the iron loss is furnished electrically as part 
of the input into the stator. At speeds above synchronism this 
power is supplied through the rotor from the prime mover. 

(e) In a direct-current ammeter, if it has a piece of iron as 
its moving element, residual magnetism in this iron causes inac- 
curacies in its indications. With the same current the indication 
of the instrument is smaller when the current is increasing than 
when it is decreasing; this can be imderstood with reference to 



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44 THE MAGNETIC CIRCUIT [Art. 19 

the hysteresis loop. With alternating current the effect of hystere- 
sis is automatically eliminated by the reversals of the current 
which passes through the instrument. 

From these examples the reader can judge as to the effect of 
hysteresis in other types of electrical apparatus not considered 
above. 

Prob. 11. Show that in an 8-pole direct-current motor running at a 
speed of 525 r.p.m. the armature core and teeth undergo 35 complete 
hysteresis cycles per second. 

Prob. 12. Show that for two points in an armature stamping, taken 
on the same radius, one in a tooth, the other near the inner periphery of 
the armature, the hysteresis loops are displaced in time by one-quarter 
of a cycle. 

19. The Total Core Loss. In practical calculations on electrical 
machinery the total core loss is of interest, rather than the hystere- 
sis and the eddy current losses separately. For such computations 
empirical curves are used, obtained from tests on steel of the same 
quality and thickness. The curves of total core loss given in 
Fig. 10 have been compiled from various sources, and give a 
fair idea of the order of magnitude of core loss in various grades 
of commercial steel laminations. The specimens were tested in 
the Epstein apparatus, which is a miniature transformer (see 
the author's Experimental Electrical Engineering ^ Vol. 1, p. 197), 
and the values given can be used for estimating the core loss in 
transformers and in other stationary apparatus with a simple 
magnetic circuit. 

In using the curves one should note that the ordinates are watts 
per cubic decimeter of laminations, hence the gross volimae and 
not the volimae of the iron itself is represented. On the other hand, 
the abscissae are the true flux densities in the iron. In choosing a 
material the following points are worthy of note: (1) Silicon steel 
is now used for 60-cycle transformers, almost to the exclusion of 
any other, on accoimt of its lower core loss; it is sometimes used 
for 25-cycle transformers also. (2) The material called " Good 
carbon steel " is that which is used for induction motor stators, 
and in general for the armatures of alternating and direct- 
current machinery ; also, sometimes for the cores of -low frequency 
transformers. (3) The material called " Ordinary carbon steel " 
should be used only in those cases for which the core loss is of 
small importance. 



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Chap. Ill] 



HYSTERESIS AND EDDY CURRENTS 



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46 THE MAGNETIC CIRCUIT [Akt. 19 

The thickness of lamination to be used in each particular 
case is a matter of judgment based on previous experience, and 
no general rule can be laid down, except what is said in Art. 17 
above, in regard to the factors upon which the eddy-current loss 
depends. The gauges 26 to 29 are representative of the usual 
practice. If it should be necessary to estimate the core loss for a • 
different thickness and at another frequency than those given in 
Fig. 10, the method explained in Art. 22 below may be used. 

The core loss in the armatures and teeth of revolving machinery 
is found from tests to be considerably above that calculated from 
the curves of loss on the same material when tested in stationary 
strips. This is probably due in part to the fact that the conditions 
of magnetization are different in the two cases. In the one case 
the cycles of magnetization are due to a 'pulsating m.m.f ., which 
simply changes its magnitude; in the other case to a gliding m.m.f., 
with which the magnetic intensity at a point changes its direction 
as well. Besides, the distribution of flux densities in teeth and in 
armature cores is very far from being uniform. Therefore, 
when using the curves given in Fig. 10, for the calculation of iron 
loss in generators and motors, it is necessary to multiply the results 
by certain empirical coefficients obtained from the results of tests 
made on similar machines. Mr. I. E. Hanssen recommends add- 
ing 30, 35, and 40 per cent to the loss calculated from the curves 
obtained on stationary samples when estimating the iron loss in 
an armature back of its teeth, at 25, 40, and 60 cycles respectively. 
For teeth he recommends adding 30, 60, and 80 per cent, at the 
same frequencies.^ These values are quoted here merely to give 
a general idea of the magnitude of the excess of core loss in revolv- 
ing machinery; a responsible designer should compile the values 
of such coefficients from actual tests made on the particular class 
of machines which he is designing. 

Some engineers do not use for revolving machinery values of 
core loss obtained on stationary samples, but plot the curves of core 
loss obtained directly from tests on machines of a particular kind, 
for various frequencies and flux densities. This is a reliable and 
convenient method provided that sufficient data are available to 
separate the core loss in the teeth from that in the core itself. Mr. 
H. M. Hobart advocates this method, and curves of core loss 

* Hanssen, " Calculation of Iron Losses in Dynamo-electric Machinery," 
Trans. Amer. Inst. Elec. Eng., Vol. 28 (1909), Part II, p. 993. 



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Chap. Ill] HYSTERESIS AND EDDY CURRENTS 47 

obtained directly from actual machines will be found in his several 
books on electric machine design. 

It is customary now to characterize a lot of steel laminations 
with respect to its core loss by the co-csAled figure of loss (Verlust- 
zififer), which is the total core loss in watts per irnit of weight, at a 
standard frequency and flux density. In Europe the figure of loss 
is imderstood to be the watts loss per kilogram of laminations, at 
50 cycles and at a flux density of 10 kilolines per square centimeter; 
the test to be performed in an Epstein apparatus imder definitely 
prescribed conditions.^ Sometimes a second figure of loss is 
required, referring to a density of 15 kilolines per square centi- 
meter, when the laminations are to be used at high flux densities. 
In this country a figure of loss is sometimes used which gives the 
watts loss per pound of material at 60 cycles and at a flux density 
of 60 kilolines per square inch (or else at 10 kilolines per square 
centimeter; see the paper mentioned in problem 20 below). 

In some cases it is required to estimate the hysteresis and the 
eddy current losses separately ; also it is sometimes necessary to 
separate the two losses knowing a curve of the total loss. These 
calculations arp explained in the articles that follow. 

Prob. 13. The core of a 60-cycle transformer weighs 89 kg.; the 
gross cross-section of the core is 8 by 10 cm., of which 10 per cent is taken 
by the insulation between the laminations. The total flux alternates 
between the values of ±0.49 megaline. If the core is made of gauge 26 
good carbon steel, what is the total core loss according to the curves in 
Fig. 10? Solution : The flux density is 490/(8 X 10 X 0.9) =6.8 kl/sq. cm. 
The core loss per cubic decimeter at this density and at 60 cycles is, 
according to the curve, equal 11.5 watt. The volume of the laminations, 
including the insulation, is 89/7 = 12.7 cu. dm. The total loss is 11.5 X 
12.7 = 146 watts. Ans. 146 watts. 

Prob. 14. What flux density could be used in the preceding problem 
if the core were made of silicon-steel laminations, gauge 29, provided that 
the total core loss be kept the same in both cases? 

Ans. About 9 kl/sq. cm. 

Prob. 16. Calculate the core loss in the stationary armature of a 60- 
cycle 450-r.p.m. alternator of the following dimensions: bore 180 cm.; 
gross axial length 24 cm.; two air-ducts 0.8 cm. each; radial width of 
stampings back of the teeth, 15 cm.; the machine has 144 slots, 2 cm. 
wide and 4.5 cm. deep. The core is made of 26-gauge good carbon steel; 
the useful flux per pole is 4.65 megalines, and two-thirds of the total num- 
ber of teeth carry the flux simultaneously. Use Mr. Hanssen's coefficients. 

1 See Elektrotechnische ZeUschrift, Vol. 24 (1903), p. 684. 



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48 THE MAGNETIC CIRCUIT [Akt. 20 

Note: All parts of the core and all the teeth are subjected to complete 
cycles of magnetization in succession ; therefore, in calculating the core 
loss the total volume of the core and of the teeth must be multiphed by 
the loss per cubic decimeter, corresponding to the maximum magnetic 
density in each part. The density in a tooth varies along its length, 
being a maximum at the tip. The average density may be assumed to be 
equal to that at the middle of the teeth. Ans. About 9 kw. 

20. Practical Data on Hysteresis Loss. The energy lost in 
hysteresis per cycle per kilogram of a given material depends only 
upon the maximum values of B and H, and does not depend upon 
the manner in which the magnetizing current is varied with the time 
between its positive and negative maxima. It is only at very high 
frequencies, such as are used in wireless telegraphy, that the par- 
ticles of iron do not seem to be able to follow in their grouping the 
corresponding changes in the exciting current. With such high 
frequencies iron cores are not only useless, but positively harmful. 
However, at ordinary commercial frequencies the loss of power 
Ph due to hysteresis is proportional to the number of cycles per 
second and can be expressed as 

P^=/.7.F(5) watt., 

where /is the number of magnetic cycles per second, F is the vol- 
ume of the iron, and F{B) is a function of the maximum flux den- 
sity B, F{B) represents the loss per cycle per cubic imit of 
material, and is therefore equal to the area of the hysteresis loop 
in Fig. 7. 

One can assume empirically that the unit loss per cycle, F(jB), 
increases as a certain power n of J5, this power to be determined 
from tests. The preceding formula becomes then 

Ph=T}fVB''wB.tt., (20) 

where t) is an empirical coefficient which depends upon the quality 
of the iron and upon the imits used. Dr. Steinmetz foimd from 
numerous experiments that the exponent n varies between 1.5 and 
1.7, and proposed for practical use the formula 

Pfc=>?/F5^-^X 10-7 watt, .... (21) 

where the factor IQ-^ is introduced in order to obtain convenient 
values for t) when B is in maxwells per square centimeter, and V 
is in cubic centimeters. It is more convenient for practical calcula- 



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Chap. Ill] HYSTERESIS AND EDDY CURRENTS 49 

tions to use B in kilolines per square centimeter, and V in cubic 
decimeters. In this case the constant 10"'^ is not necessary (see 
Prob. 17 below) ; but the student must now remember to multiply 
by 6.31 the values of rj foimd in the various pocketbooks. 

Hysteresis loss cannot be represented always with sufficient 
accuracy by formula (21) or (20) over a wide range of values of 5, 
because the exponent n itself seems to increase with B. Where 
greater accuracy is required at medium and high flux densities the 
following formula may be used: 

PH=mr)'B + rB^) (21a) 

In this formula the term with B^ automatically becomes of more 
and more importance as B increases. By selecting proper values 
for 7j^ and tj^^ a given experimental curve of 'loss can be approxi- 
mated more closely than by means of formula (21). On the other 
hand, formula (21) is more convenient for comparison and analysis. 

Curves of hysteresis loss and values of the constant tj will be 
foimd in various handbooks and pocketbooks. It is hardly worth 
while giving them here, because hysteresis loss varies greatly with 
the quality of iron and with the treatment it is given before use. 
Moreover, the quality of the iron used in electrical machinery is 
being improved all the time, so that a value of t) given now may 
be too large a few years from now. 

Considerable effort is being constantly made to improve the 
quality of the iron used in electrical machinery so as to reduce its 
hysteresis loss. The latest achievement in this respect is the pro- 
duction of the so-called silicon sted, also called alloyed sted, which 
contains from 2.5 to 4 per cent of silicoxi. This steel shows a much 
lower hysteresis loss than ordinary carbon steel. Incidentally, 
the electric resistivity of silicon steel is about three times higher 
than that of ordinary steel, so that the eddy -current loss is reduced 
about three times. The advantage that silicon steel has over car- 
bon steel is clearly seen in Fig. 10. Silicon steel is largely used 
for transformer cores because it permits the use of higher flux 
densities, and therefore the reduction of the weight and cost of a 
transformer, in spite of the fact that silicon steel itself costs more 
per kilogram than carbon steel. 

Another great advantage of silicon steel is that it is practically 
non-aging', this means that the hysteresis loss does not increase 
with time. An increase in the hysteresis loss of a transformer 



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50 THE MAGNETIC CIRCUIT [Art. 20 

during the first few years of its operation used to be a serious 
matter in the design and operation of transformers, because of the 
subsequent overheating of the core and of the coils. Silicon steel 
shows practically no increase in its hysteresis loss after several 
years of operation. Moderate heating, which considerably in- 
creases the hysteresis loss in ordinary steel, has no effect on silicon 
steel. 

Impurities which are of such a nature as to produce a softer 
iron or steel and a material of higher permeability, are as a rule 
favorable to the reduction of the hysteresis loss, and vice versa. 
Mechanical treatment and heating are also very important in their 
effects on hysteresis loss. In particular, pimching and hammering 
increases hysteresis loss, while annealing reduces it. Therefore 
laminations are always annealed carefully after being pimched 
into their final shape. 

The requirements for the steel used in permanent magnets are 
entirely different from those for the cores of electrical machinery. 
In permanent magnets a large and wide hysteresis loop is desired, 
because it means a high percentage of residual magnetism (ratio 
of CO to AP, Fig. 7) and a large coercive force, OF. Both are 
favorable for obtaining strong permanent magnets of lasting 
strength. Combined carbon is particularly important for obtain- 
ing these qualities, as is also the proper heat treatment after mag- 
netization. 

Prob. 16. In the 60-cycle transformer given in prob. 13, the core 
weighs 89 kg. and is made of 26 gauge good carbon steel. The maxi- 
mum flux density is 6.8 kl./sq. cm. What is the hysteresis loss assuming 
1? to be equal to 0.0012? Ans. About 124 watt. 

Prob. 17. What is the constant in formula (21) in place of 10-^, if, 
with the same i?, the density B is in kilo-maxwells per sq. cm., and the 
volume is in cubic decimeters? Ans. 6.31. 

Prob. 18. Show how to determine the values of i? and n in eq. (20), 
knowing the values Wi and W2 of the energy lost per cycle at two given 
values of maximum flux density, Bi and B2. 

Ans. n = (log TTj-log TrO/(log Br-log B^). 

Prob. 19. The following values of hysteresis loss per cu. decimeter 
have been determined from a test at 25 cycles (after eliminating the eddy 
current loss) : 

Flux density in kl/sq.cm., B = 5.0 |6.5 [8.0 110.0 
Hysteresis loss in watts, Ph = lSO I 2.00 | 2.88 | 4.11 
What are the values of tj and n in formula (20)? Suggestion; 
Use logarithmic paper to determine the most probable value of n, by 



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Chap. Ill] HYSTERESIS AND EDDY CURRENTS 51 

drawing the straight line log Ph=ri log B+log Const. See the author's 
Experimental Electrical Engineering, Vol., 1, p. 202. 

Ans. Pa = 0.00368 /FB»-« 

21. Eddy Ctirrent Loss in Iron. With the thin laminations 
used in the cores of electrical machinery the eddy -current loss in 
watts can be represented by the formula 

Pe=eV{tfB)^ (22) 

where s is a constant which depends upon the electrical resistivity 
of the iron, its temperature, the distribution of the flux, the wave 
form of the exciting current, and the units used. V is the volume 
or the weight of the core for which the loss is to be computed; t is 
the thickness of laminations, / the frequency of the supply, and B 
the maximum flux density during a cycle. If B is different at 
different places in the same core, the average of these should 
be taken; {B is the time maximum but the space average). 
Sometimes formula (22) contains also 10 to some negative power 
in order to obtain a convenient value of e. 

Formula (22) can be proved as follows: The loss of power in a 
lamination can be represented as a sum of the i^r losses for the 
small filaments of eddy current in it. But v^r=e^/r; it can be 
shown that the expression in parentheses in formula (22) is pro- 
portional to the sum of e^/r per imit volume. When the frequency 
/ increases say n times, the rate of change of the flux, d^/dt, and 
consequently the e.m.fs. induced in the iron are also increased n 
times. Therefore, the loss which is proportional to e^ increases n^ 
times. In other words, the loss is proportional to the square of 
the frequency. Similarly, the induced voltage is proportional to 
tlie flux density 5; and consequently, the loss is proportional to B^. 

To prove that the loss is proportional to the square of the 
thickness of laminations one must remember that increasing the 
thickness n times increases the flux and the induced e.m.f . within 
any filament of eddy current also n times. But the resistance of 
each path is reduced n times (neglecting the short sides of the rect- 
angle). Consequently, the expression e^/r is increased n^ times. 
However, inasmuch as the volume of the lamination is also 
creased n times, the loss per unit volume is only n^ times larger. In 
other words, the loss per unit volume increases as t^. A more 
rigid proof of this proposition is given in problem 21 below. 



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52 THE MAGNETIC CIRCUIT [Art. 22 

For values of e the reader is referred to pocketbooks; the 
numerical values given there must, however, be used cautiously, 
because the eddy-current loss depends on some factors such as the 
care exercised in assembling, and the actual distribution of the 
flux, which factors can hardly be taken into account in a formula. 
As a matter of fact, formula (22) is used now less and less in prac- 
tical calculations, the engineer relying more upon experimental 
curves of total core loss (Fig. 10). 

Prob. 20. According to the experiments of Lloyd and Fischer [Trans- 
Amer. Inst, Eke, Engs., Vol. 28 (1909), p. 465] the eddy-current loss in 
silicon-steel laminations of gauge 29 (0.357 mm. thick) is from 0.12 to 0.18 
watts per poimd at 60 cycles and at 5 = 10,000 maxwells per sq. cm. What 
is the value of the coefficient e in formula (22) if P is in microwatts, V is 
the weight of the core in kg. (not the volume, as before) ; if also < is in nmi., 
and ^ is in kilolines per sq. cm.? 

Ans. From 5.78 to 8.67; 7.2 is a good practical average. 

Prob. 21. Prove that the loss of power caused by eddy currents, per 
unit volume of thin laminations, is proportional to the square of the thick- 
ness of the laminations. Solution : The thickness t of the sheet (Fig. 9) 
being by assumption very small as compared with its width a, the paths 
of the eddy ciurent may be considered to be rectangles of the length a and 
of different widths, ranging from t to zero. Consider one of the tubes of 
flow of current, of a width 2a;, thickness dx, and length I in the direction 
of the lines of magnetic force. Let the flux density vary with the time 
between the limits ±B, Then the maximum flux linking with the tube 
of current under consideration is approximately equal to 2axB; therefore, 
the effective value of the voltage induced in the tube can be written in the 
form e^CaxBff where C is a constant, the value of which we are not con- 
cerned with here. The resistance of the tube is p{2a +4x) /(Idx), or very 
nearly 2ap/{ldx). Thus we have that the iV loss, or the value oie^/r 
for the tube imder consideration, is dPe^C^ax^B^pidx/2p, Integrat- 
ing this expression between the limits and tl2 we get Pe= C^af^B^pi/ 4Sp, 
But the volume of the lamination is 7 =a<Z. Dividing P by 7 we find 
that the loss per unit volume is proportional to (tfBy} 

Prob. 22. Prove that the loss of power by eddy ciurents per unit 
volume in roimd iron wires is proportional to the square of the diameter of 
the wire. The flux is supposed to pulsate in the direction of the axes of 
the wires, and the lines of flow of the eddy ciurents are concentric circles 
Hint : Use the method employed in the preceding problem. 

22. The Separation of Hysteresis from Eddy Currents. It is 
sometimes required to estimate the total core loss for a thickness 

*For a complete solution of this and the following problem, including 
the numerical values of C, see Steiometz, AUemaiing Current Phenomena 
(1908), Chap. XIV. 



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Chap. Ill] HYSTERESIS AND EDDY CURRENTS 53 

of steel laminations other than those given in Fig. 10, or at a diflFer- 
ent frequency. For this purpose, it is necessary to separate the 
loss due to hysteresis from that due to eddy currents, because the 
two losses follow diflferent laws, expressed by eqs. (20) and (22) 
respectively. 

In order to separate these losses at a certain flux density it is 
necessary to know the value of the total core loss at this density, 
and at two different frequencies. For a given sample of lamina- 
tions, the total core loss P at a constant flux density and at a 
variable frequency /, can be represented in the form 

P=Hf+Fp, (23a) 

where Hf represents the hysteresis loss, and Fp the eddy or Fou- 
cault current loss. H is the hysteresis loss per cycle, and F is the 
eddy-current loss when / is equal to one cycle per second. Writing 
this equation for two known frequencies, two simultaneous equa- 
tions are obtained for H and F, from which H and F can be deter- 
mined. 

In practice the preceding equation is usually divided by/, be- 
cause in the form 

P/f^H^Ff, (236) 

it represents the equation of a straight line between P/f and /. 
This form is particularly convenient when the values of P are 
known for more than two frequencies. In this case the values of 
P/f are plotted against / as abscissae, and the most probable 
straight line is drawn through the points thus obtained. The 
intersection of this straight line with the axis of ordinates gives 
directly the value of H. After this, F is found from eq. (236). 

Knowing H and F at a certain flux density, the separate losses 
Hf and Fp can be calculated for any desired frequency. For the 
same material, but of a different thickness, the hysteresis loss per 
kilogram weight is the same, while the eddy-current constant F 
varies as the square of the thickness, according to eq. (22). Thus, 
knowing the eddy loss at one thickness it can be estimated for any 
other thickness. 

It is sometimes required to estimate the iron loss at a flux den- 
sity higher than the range of the available curves; in other words, 
the problem is sometimes put to extrapolate a curve like one of 
those in Fig. 10. There are two cases to be considered. 



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54 THE MAGNETIC CIRCUIT [Art. 22 

(A) If two or more curves for the same material are available, 
taken at different frequencies, the hysteresis is first separated from 
the eddy-current loss as is explained before, for several flux densi- 
ties within the range of the curves. Then the exponent, according 
to which the hysteresis loss varies with the flux density is found, 
by plotting the hysteresis loss to a logarithmic scale (see problems 
18 and 19 above). Finally the two losses are extrapolated. In 
extrapolating, the hysteresis loss is assumed to vary according to 
the same law, and the eddy current loss is assumed to vary as the 
square of the flux density; see eq. (22). 

(B) Should only one curve of the total loss be available for 
extrapolation, this curve may be assumed to be a parabola of the 
form P—aB+bB^, Dividing the equation throughout by B we 
get 

P/B=a+bB (24) 

This is the equation of a straight line between P/B and B. Plot- 
ting P/B against the values of B as abscissae, a straight line is 
obtained which can be easily extrapolated. In some cases the 
values of P/B thus plotted give a line with a perceptible curva- 
ture. Nevertheless, the curvature is much smaller than that of 
the original P curve, so that the P/B curve can be extrapolated 
with more certainty, especially if the lower points be disregarded.^ 

Prob. 23. From the curves in Fig. 10 calculate the core loss per cubic 
decimeter of 29-gauge silicon-steel laminations, at a flux density of 10 
kl/sq.cm. and at 40 cycles. Ans. About 10 watts. 

Prob. 24. Using the data obtained in the solution of the preceding 
problem calculate the figure of loss of 26-gauge laminations at 60 cycles. 

Ans. 2.7 watt/kg. 

Prob. 26. Check the curve of total core loss for the ordinary carbon 
steel at 40 cycles with the curves for 25 and 60 cycles. 

Prob. 26. Extrapolate the curve of core loss for the silicon steel at 25 
cycles up to the density of 20 kl/sq.cm. Which method is the more 
preferable? Ans. 22 watts per cu.dm. at B = 20, 

Prob. 27. Show that the core loss curve for ordinary carbon steel, at 
60 cycles, follows closely eq. (24). 

1 If the P/B curve should prove to be a straight line, then it is probable 
that the hysteresis loss follows eq. 21a more nearly than eq. 20. In this case, 
even if we had data for two frequencies, method (B) would be both more 
accurate, and more simple than method (A). 



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CHAPTER IV 
INDUCED E.M.F. IN ELECTRICAL MACHINERY 

23. Methods of Inducing E.M.F. The following are the prin- 
cipal cases of induced e.m.f. in electrical machinery and apparatus: 

(a) In a transformer, an alternating magnetizing current in the 
primary winding produces an alternating flux which links with 
both windings and induces in them alternating e.mis. A similar 
case is that of a variable current in a transmission line which 
induces a voltage in a telephone line which runs parallel to it. 

(6) In a direct-current machine, in a rotary converter, and in 
a homopolar machine electromotive forces are induced in the 
armature conductors by moving them across a stationary magnetic 
field. 

(c) In an alternator and in a synchronous motor, with a sta- 
tionary armature and a revolving field, electromotive forces are 
induced by making the magnetic flux travel past the armature 
conductors. 

(d) In a polyphase induction motor the currents in the stator 
and in the rotor produce together a resultant magnetomotive force 
which moves along the air-gap and excites a gliding (revolving) 
flux. This flux induces voltages in both the primary and the sec- 
ondary windings. 

(e) In a single-phase motor, with or without a commutator, the 
e.m.fs. induced in the armature are partly due to the " transformer 
action," as under (a), and partly to the ''generator action," as 
under (6). 

(/) In an inductor-type alternator both the exciting and the 
armature windings are stationary ; the pole pieces alone revolve. 
The flux linked with the armature coils varies periodically, due to 
the varying reluctance of the magnetic circuit, because of the 
motion of the pole pieces. This varying flux induces an alternating 
e.m.f. in the armature winding. Or else, one may say that the 

55 



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66 THE MAGNETIC CIRCUIT [Art. 23 

fluxtravels along the air-gap with the projecting poles, and cuts 
the armature conductors. 

(g) Whenever the current varies in a conductor, e.m.fs. are 
induced not only in surrounding conductors but also in the con- 
ductor itself. This e.m.f. is called the e.m.f. of self-induction. 
Such e.m.fs. are present in alternating-current transmission lines, 
in the armature windings of alternating-current machinery, etc. 
While the e.m.f. of self-induction does not differ fimdamentally 
from the transformer action mentioned above, its practical aspect 
is such as to make a somewhat different treatment desirable. 
Inductance and its effects are therefore considered separately in 
chapters X to XII below. 

All of the foregoing cases can be reduced to the following two 
fundamental modes of action of a magnetic flux upon an electrical 
conductor: 

(1) The exciting magnetomotive force and the winding in which 
an e.m.f. is to be induced are both stationary, relatively to one 
another; in this case the voltage is induced by a varying magnetic 
flux. Changes in the flux are produced by varying either the 
magnitude of the m.m.f ., or the reluctance of the magnetic circuit. 
This method of inducing an e.m.f. is usually called the transformer 
action. 

(2) The exciting magnetomotive force and the winding in which 
the e.m.f. is to be induced are made to move relatively to each 
other, so that the armature conductors cut across the lines of the 
flux. This method of inducing an e.m.f . is conventionally referred 
to as the generator action. 

By analyzing the transformer action more closely it can be 
reduced to the generator action, that is to say to the " cutting " of 
the secondary conductor by lines of magnetic flux or force. This is 
so, because in reality the magnetic disturbance spreads out in all 
directions from the exciting winding, and when the current in the 
exciting winding varies the magnetic disturbance travels to or 
from the winding in directions perpendicular to the lines of force 
(Fig. 11). This traveling flux cuts the secondary conductor and 
induces in it an e.m.f. However, the question as to whether an 
e.m.f. is induced by a change in the total flux within a loop, or by 
the cutting of a conductor by a magnetic flux is still in a somewhat 
controversial state ;i although Bering's experiment is a strong 

* Carl Hering, " An Imperfection in the Usual Statement of the Funda- 



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Chap. IV] 



INDUCED E.M.F. 



67 



argument in favor of the theory of " cutting " of lines of force. 
He showed that no e.m.f. is induced in an electric circuit when a 
flux is brought in or out of it without actually cutting any of the 
conductors of the electric circuit. For practical purposes it is 
convenient to distinguish the transformer action from the genera- 
tor action, so that the matter of imifying the statements (1) and 
(2) into one more general law is of no immediate importance. 

24. The Formtdae for Induced E.M.F. In accordance with the 
definition of the weber given in Art. 3, we have 

e=^dO/dt, (25) 

where e is the instantaneous e.m.f . in volts, induced by the trans- 




FiG. 11. — ^E.M.F. induced by transformer action. 

former action in a turn of wire which at the time t is linked with a 
flux of <P webers. The value of e is determined not by the value of 
but by the rate at which varies with the time. In the case of the 
generator action dO in formula (25) represents the flux which the 
conductor under consideration cuts during the interval of time dt. 
It can be shown that the two interpretations of d0 lead to the 
same result. Namely , in the case of the transformer action (Fig. 1 1) , 
the new flux, d(P, is brought within the secondary turn by cutting 
through the conductor of this turn. Therefore, in the case of the 

mental Law of Electromagnetic Induction, Trans. Amer. Inst. Elec. Engs.f 
Vol. 27 (1908), Part. 2, p. 1341. -Fritz Emde, Das Induktionsgesetz, Elek- 
trotechnik und MaschinenbaUy Vol. 26 (1908); Zum Induktionsgesetz, ibid., 
Vol. 27 (1909); De Baillehache, Sur la Loi de I'lnduction, BuU. SocieU Inter- 
natumcde des Electriciens, Vol. 10 (1910), pp. 89 and 288. 



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58 THE MAGNETIC CIRCUIT [Art. 24 

transf onner action d* can also be considered as the flux which cuts 
the loop during the time dt, the same as in the generator action. 
On the other hand, the moving conductor in a generator is a part 
of a turn of wire, and any flux which it cuts either increases or 
decreases the total flux linking with the loop. Consequently, in the 
case of the generator action d0 can be interpreted as the change of 
flux within the loop, the same as in the transformer action. Thus, 
the mathematical expression for the induced e.m.f. is the same 
in both cases, provided that the proper interpretation is given to 
the value of d0. 

The sign minus in formula (25) is understood with reference to 
the right-hand screw rule (Art. 1), i.e., with reference to the direc- 
tion of the current which would flow as a result of the induced 
electromotive force. Namely, the law of the conservation of 
energy requires that this induced current must oppose any change 
in the flux linking with the secondary circuit. If this were other- 
wise, a slight increase in the flux would result in a further indefinite 
increase in the flux, and any slight motion of a conductor across a 
magnetic field would help further motion. 

The positive direction of the induced e.m.f . is understood to be 
that of the primary current which excites the flux at the moment 
under consideration. If the flux linked with the secondary circuit 
increases, d0/dt in formula (25) is positive, but the secondary 
current must be opposite to the primary in order to oppose the 
increase. Thus, the secondary current is negative, and by assump- 
tion the induced e.m.f. e is also negative. Therefore, the sign 
minus is necessary in the formula. When the flux decreases, 
d0/dt is negative, but the secondary current is positive, because it 
must oppose the reduction in flux. Hence, in order to make e a 
positive quantity, the sign minus is again necessary. 

The following two special cases of formula (25) are convenient 
in applications. Formula (25) gives the instantaneous value of 
the induced e.m.f.; it is once and a while required to know the 
average e.m.f. induced during a finite change of the flux from 0i 
to 02' By definition, the average e.m.f. is 



fe — hJti 



it 

edt, 



where tx is the initial moment and fe the final moment of the 
interval of time during which the change in the flux takes place. 



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Chap. IV] INDUCED E.M.F. 59 

Substituting in this equation the value of e from (25), and integ- 
rating, we get 

eat«=(*i-<^2)/fe-*i). ..... (26) 

This shows that the average value of an induced e.m.f . does 
not depend upon the law according to which the flux changes with 
the time, and is simply proportional to the average rate of change 
of the flux. 

As another special form of eq. (25) consider a straight con- 
ductor of a length I centimeters moving at a velocity of v centi- 
meters per second across a uniform magnetic field of a density of B 
webers per sq. cm. Let 5, Z, and v be in three mutually perpendic- 
ular directions. The flux d0 cut by the conductor during an infini- 
tesimal element of time di is equal to BlvdU Substituting this 
value into eq. (25) we get, apart from the sign minus, 

e=^Blv (27) 

Should the three directions, 5, Z, and r, be not perpendicular to 
each other, I in eq. (27) is understood to mean the projection of 
the actual length of the conductor, perpendicular to the field, and 

V is the component of the velocity normal to B and I, Both B and 

V may vary with the position of the conductor, in which case eq. 
(27) gives the value of the instantaneous voltage. If, at a cer- 
tain moment, the various parts of the conductor cut across a field 
of different density, eq. (27) must be written for an infinitesimal 
length of the conductor, thus: de=Bv'dl, and integrated over 
the whole length of the conductor; 

Besides the rule given above, the direction of the e.m.f . induced 
by the generator action can also be determined by the familiar 
three-finger rule, due to Fleming, and given in handbooks and ele- 
mentary books on electricity. This rule is useful beause it empha- 
sizes the three mutually perpendicular directions, those of the 
flux, the conductor, and the relative motion. In applying this 
rule to a machine with a stationary armature one must remember 
that the direction of the motion in Fleming's rule is that of the 
conductor, and therefore is opposite to the direction of the actual 
motion of the magnetic field. 

Problem 1. A secondary winding is placed on the ring (Fig. 1) and is 
connected to a ballistic galvanometer. lict the number of turns in the 
secondary winding be n, the flux linking with each turn be webers, and 



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60 THE MAGNETIC CIRCUIT [Art. 25 

the total resistance of the secondary circuit be r ohms. Show that when 
the current in the primary winding is reversed, the discharge through 
the galvanometer is equal to 2n<P coulombs. 

Prob. 2. A telephone line runs parallel to a direct-current trolley 
feeder for 20 kilometers. When a current of 100 amperes flows through 
the feeder a flux of 2 kilo-maxwells threads through the telephone loop, 
per meter of its length. What is the average voltage induced in the tele- 
phone line when the current in the trolley feeder drops from 600 to 50 amp. 
within 0.1 sec? Ans. 22 volts. 

Prob. 3. Determine the number of armatxire conductors in series in a 
550 volt homopolar generator of the axial type, running at a peripheral 
speed of about 100 meters per sec., when the length of the armature iron 
is 50 centimeters, and the flux density in the air-gap is between 18 and 19 
kilolines per sq. cm. Note: For the construction of the machine sefe the 
Standard Handbook, index, imder " Generators, homopolar." 

Ans. 6. 

Prob. 4. Draw schematically the armature and the field windings of a 
shimt-wound direct-current generator, select a direction of rotation, and 
show how to connect the field leads to the brushes so that the machine 
will excite itself in the proper direction, 

Prob. 6. From a given drawing of a direct-current motor predict its 
direction of rotation. 

Prob. 6. In an interpole machine the average reactance voltage per 
commutator segment during the reversal of the current is calculated to be 
equal to 34 volts. What is the required net axial length of the commu- 
tating pole to compensate for this voltage if the peripheral speed of the 
machine is 65 meters per second, and the flux density under the pole is 
6 kl. sq.cm.? The armature winding has two turns per commutator seg- 
ment. Ans. 22 cm. 

25. The Induced E.M.F. in a Transformer. The three types 
of transformers used in practice are shown in Figs. 12, 13, and 14. 
Considering the iron core as a magnetic link, and a set of primary 
and secondary coils as an electric link, one may say that the core- 
type transformer has one magnetic link and two electric links; 
the shell-type has one electric link and two magnetic links; the 
combination or cruciform type has one electric and four magnetic 
links. Still another type, not used in practice, can be obtained 
from the core-type by adding two or more electric links to the same 
magnetic link. Each electric link is imderstood to consist of two 
windings: the primary and the secondary. 

When the primary winding is connected to a source of constant- 
potential alternating voltage and the secondary winding is con- 
nected to a load, alternating currents flow in both windings and an 
alternating magnetic flux is established in the iron core. If the 



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Chap. IV] 



INDUCED E.M.F. 



61 



primary electric circuit, that is, the one connected to the source of 
power, were perfect, that is, if it possessed no resistance and no 
reactance, the alternating magnetic flux in the core would be the 
same at all loads. It would have such a magnitude that at any 
instant the coimter-e.m.f. induced by it in the primary wind- 
ing would be practically equal and opposite to the impressed 
voltage. In reality the resistance and the leakage reactance of 
ordinary commercial transformers are so low that for the purposes 
of calculating the magnetic circuit the primary impedance drop may 
be disregarded, and the mag- 
netic flux considered constant 
and independent of the load. 

If the primary applied volt- 
age varies according to the sine 
law, which condition is nearly 
fulfilled in ordinary cases, the 
counter-e.m.f ., which is practi- 
cally equal and opposite to it, 
also follows the same law. 
Hence, according to eq. (25), 
the magnetic flux must vary 
according to the cosine law, 
because the derivative of the 
cosine is minus the sine. In 
other words, both the flux and 
the induced e.m.f. vary accord- 
ing to the sine law, but the two 

sine waves are in time quadrature with each other. When the 
flux reaches its maximum its rate of change is zero, and therefore 
the coimter-e.m.f . is zero. When the flux passes through zero its 
rate of change with the time is a maximum, and therefore the 
induced voltage at this instant is a maximum. 

Let<P„jbe the maximum value of the flux in the core, in webers, 
and let / be the frequency ^ of the supply in cycles per second. 
Then the flux at any instant t is <l> = (P„iCOs 2nftj and the e.m.f. 
induced at this moment, per turn of the primary or secondary 
winding is 

e= -dO/dt=2nf0,n sin 2nft. 
Thus, the maximum value of the induced voltage per turn is 




Fig. 12. — A core-type transformer. 



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62 



THE MAGNETIC CIRCUIT 



[Art. 25 



2TzfiPrn\ hence, the effective value is 2;r/(^ni/v'2 = 4.44/(1)^. Let 
there be iVi primary turns in series; the total primary voltage is 
then equal to iVi times the preceding value. Expressing the flux 
in megalines we therefore obtain the following practical formula 
for the induced voltage in a transformer: 



i;i = 4.44/iVi(p^lO-^ 



(28) 



Core^ 




CoUs 



Fig. 13.— a shell-type 
transformer. 



Mica 



Fig. 14. — A cruciform-type 
transformer. 



In practice, E^ is assumed to be equal and opposite to the applied 
voltage (for calculating the flux only, but not for determining the 
voltage regulation of the transformer). Formula (28) holds also 
for the secondary induced voltage E2 if the number of secondary 
turns in series ^2 be substituted for iVi. The voltage per turn is 
the same in the primary and in the secondary winding; therefore, 
the ratio of the induced voltages is equal to that of the number 



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Chap. IV] INDUCED E.M.F. 63 

of turns in the primary and secondary windings: that is, we 
have Ei:E2=Ni: N2. 

Prob. 7. A,60-cycle transformer is to be designed so as to have a flux 
density in the core of about 9 kl./sq.cm.; the difference of potential 
between consecutive turns must not exceed 5 volts. What is the required 
cross-section of the iron? Ans. 210 sq.cm. 

Prob. 8. The transformer in the preceding problem is to be wound for 
6600 V. primary, and 440 v. secondary. What-are the required numbers 
of turns? Ans. 1320 and 88. 

Prob. 9. Referring to the transformer in the preceding problem, what 
are the required numbers of turns if three such transformers are to be used 
Y-connected on a three-phase system, for which the line voltages are 6600 
and 440 respectively? Ans. 765 and 51. 

Prob. 10. In a 110-kilovolt, 25-cycle transformer for Y-connection the 
net cross-section of the iron is about 820 sq.cm. and the permissible maxi- 
mum flux density is about 10.7 kl/sq.cm. What is the number of turns 
in the high-tension winding? Ans. 6500. 

Prob. 11. The secondary of the transformer in the preceding problem 
is to be wound for 6600 v., delta connection, with taps for var3dng the 
secondary voltage within ±5 per cent. Specify the winding. 

Ans. 709 turns; taps taken after the 34th and 68th turn. 

Prob. 12. Explain the reason for which a 60-cycle transformer usually 
runs hot even at no load, when connected to a 25-cycle circuit of the same 
voltage. Show from the core-loss ciuv^es that the voltage must be 
reduced to from 75 to 85 per cent of its rated value in order to have the 
normal temperature rise in the transformer, at the rated current. 

Prob. 13. Show graphically that the wave of the flux, within a trans- 
former, becomes more and more peaked when the wave of the applied e.m.f. 
becomes more and more flat, and vice versa. Hint: The instantaneous 
values of e.m.f . are proportional to the values of the slope of the ciuve of 
flux. 

Prob. 14. The wave of the voltage impressed upon a transformer has 
a 15 per cent third harmonic which flattens the wave symmetrically. Show 
analytically that the corresponding flux wave has a 5 per cent third har- 
monic in such a phase position as to make the flux wave peaked. 

26. The Induced E.M.F. in an Alternator and in an Induction 
Motor. Part of a revolving field alternator is shown in Fig. 15. 
The armature core is stationary and has a winding placed in slots, 
which may be either open or half closed. The pole pieces are 
moimted on a spider and are provided with an exciting winding. 
When the spider is driven by a pime-mover the magnetic flux 
sweeps past the armature conductors and induces alternating 
voltages in them.^ In order to obtain an e.m.f. approaching a sine 

* For details concerning the different types of armature windings see the 
author's Experimental Electrical En^neering, Vol. 2, Chap. 30. 



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64 



THE MAGNETIC CIRCUIT 



[Art. 26 



wave as nearly as possible the pole shoes are shaped as shown 
in the sketch, that is to say, so as to make a variable air-gap and 
thus grade the flux density from the center of the pole to the edges. 
In high-speed turbo-alternators the field structure often has a 
smooth surface, without projecting poles (Fig. 33), in order to 
reduce the noise and the windage loss. Such a structure is also 
stronger mechanically than one with projecting poles. The grad- 
ing of the flux is secured by distributing the field winding in slots, 
so that the whole m.m.f . acts on only part of the pole pitch. 

Consider a conductor at a during the interval of time during 
which the flux moves by one pole pitch t. The average e.m.f . 




Fig. 15. — ^The cross-section of a synchronous machine. 

induced in the conductor is, according to eq. (26), equal to (^/JT, 
where is the total flux per pole in webers, and T is the time of 
one complete cycle, corresponding to 2r the space of two pole 
pitches. But !r=l//, so that the average voltage induced in a 
conductor is 



e=2/*- 



(29) 

The value of e^^^ thus does not depend upon the distribution of the 
flux in the air-gap. 

If the pole-pieces are shaped so as to give an approximately 
sinusoidal distribution of flux in the air-gap, the induced e.m.f. is 
also approximately a sine wave, and the ratio between the effect- 



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Chap. IV] INDUCED E.M.F. 65 

ive and the average values of the voltages is equal to \n/s/2 
or 1.11.^ If the shape of the induced e.m.f. departs widely from the 
sine wave the actual curve must be plotted and its form factor 
determined by one of the known methods (see the Electric Cir- 
cuit). Let the form factor in general be % ^^d let the machine 
have N armature turns in series per phase, or what is the same, 
2N conductors in series. The total induced e.m.f. in effective 
volts is then 

E=2fx02N (30) 

This formula presupposes that there is but one slot per pole per 
phase, so that the e.m.fs. induced in the separate conductors are all 
in phase with each other, and thai) their values are simply added 
together. In reality, there is usually more than one slot per pole 
per phase, for practical reasons discussed in the next article. It 
will be seen from the figure that the e.m.fs. induced in adjacent 
slots are somewhat out of phase with each other, because the crest 
of the flux reaches different slots at different times. Therefore, 
the resultant voltage of the machine is somewhat smaller than that 
according to the preceding formula. The influence of the dis- 
tribution of the winding in the slots is taken into account by mul- 
tiplying the value of £7 in the preceding formula by a coefiicient k^, 
which is smaller than imity and which is called the breadth factor. 
Introducing this factor, and assuming ;f=l.ll, which is accurate 
enough for good commercial alternators, we obtain 

£=4.44Jfc6/iV^10-2, (31) 

where is now in megalines. Values of k^ are given in the articles 
that follow. 

Formula (31) applies equally well to the polyphase induction 
motor or generator. There we also have a imiformly revolving 
flux in the air-gap, the flux density being distributed in space, 
according to the sine law. This gliding flux induces e.m.fs. in the 
stator and rotor windings. The only difference between the two 
kinds of machines is that in the synchronous alternator the field is 
made to revolve by mechanical means, while in an induction 
machine the field is excited by the polyphase currents flowing in 

* For the proportions of a pole-shoe which very nearly give a sine wave 
see Arnold, WechsdstronUechniky Vol. 3, p. 247. 



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66 THE MAGNETIC CIRCUIT [Abt. 27 

the stator and rotor windings. The formulae of this chapter also 
apply without change to the synchronous motor, because the con- 
struction and the operation of the latter are identical with those of 
an alternator; the only difference being that an alternator trans- 
forms mechanical energy into electrical energy, while a synchro- 
nous motor transforms energy in the reverse direction. In all 
cases the induced voltage is understood and not the line voltage. 
The latter may differ considerably from the former, due to the 
impedance drop in the stator winding. 

Prob. 16. A delta-connected, 2300-v., 60-cycle, 128.5-r .p.m. alternator 
is estimated to have a useful flux of about 3.9 megalines per pole. If the 
machine has one slot per pole per phase how many conductors per slot are 
needed? Ans. 8. 

Prob. 16. A 100,000-cycle alternator for wireless work has one conduc- 
tor per pole and 600 poles. The rated voltage at no load is 110 v. What 
is the flux per pole and the speed of the machine? 

Ans. 82.5 maxwells; 20,000 r.p.m. 

Prob. 17. It is desired to design a line of induction motors for a per- 
ipheral speed of 50 met. per sec, the maximum density in the air-gap to be 
about 6 kilolines per sq.cm. What will be the maximum voltage induced 
per meter of active length of the stator conductors? Hint: Use formula 
(27). Ans. 30 volt. 

Prob. 18. Formula (31) is deduced under the assumption that each 
Armature conductor is subjected to the " cutting " action of the whole 
flux. In reality, almost the whole flux passes through the teeth between 
the conductors, so that it may seem upon a superficial inspection that 
little voltage could be induced in the conductors which are embedded in 
slots. Show that such is not the case, and that the same average voltage 
is induced in the conductors placed in completely closed slots, as in the 
conductors placed on the surface of a smooth-body armature. Hint: 
When the flux moves, the same amount of magnetic disturbance must 
pass in the tangential direction through the slots as through the teeth. 

Prob. 19. Deduce eq. (31) directly from eq. (27). Can eq (31) be 
derived under the case of the transformer action? 

27. The Breadth Factor. Armature conductors are usually 
placed in more than one slot per pole per phase, for the following 
reasons : 

(a) The distribution of the magnetic field is more uniform, 
there being less bunching of the flux under the teeth; 

(b) The induced e.m.f. has a better wave form; 

(c) The leakage reactance of the winding is reduced; 

(d) The same armature punching can be used for machines with 
different numbers of poles and phases; 



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Chap.IVI induced E.M.F. 67 

(e) The mechanical arrangement and cooling of the coils is 
somewhat simplified. 

The disadvantage of a large number of slots is that more space 
is taken up by insulation, and the machine becomes more expen- 
sive, especially if it is wound for a high voltage. The electromo- 
tive force is also somewhat reduced because the voltages induced 
in different slots are somewhat out of phase with one another. 
The advantages of a distributed winding generally outweigh its dis- 
advantages, and such windings are used almost entirely. Thus, 
it is of importance to know how to calculate the value of the 
breadth factor h for a given winding. 




I Phase 1 
i ^« 2 
D " 3 

Fig. 16. — ^A fractional-pitch winding. 

In the winding shown in Fig. 15 each conductor is connected 
with another conductor situated at a distance exactly equal to the 
pole pitch. It is possible, however, to connect one armature con- 
ductor to another at a distance somewhat smaller than the pole 
pitch (Fig. 16). Such a winding is called Si fractional-pitch wind- 
ing y in distinction to the winding shown in Fig. 15; the latter 
winding is called a full-pitch or hundred-per cent pitch winding. 
It wiU be seen from Fig. 16 that, with a two-layer fractional-pitch 
winding, some slots are occupied by coils belonging to two different 
phases. The advantages of the fractional-pitch winding are: 

(a) The end-connections of the winding are shortened, so that 
there is some saving in armature copper. 

(6) The end-connections occupy less space in the axial direc- 
tion of the machine, so that the whole machine is shorter. 



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68 THE MAGNETIC CIRCUIT [Art. 28 

(c) In a two-pole or four-pole machine it is necessary to use a 
fractional-pitch winding in order to be able to place machine- 
wound coils into the slots. 

A disadvantage of the fractional-pitch winding is that the 
e.mis. induced on both sides of the same coils are not exactly in 
phase with each other, so that for a given voltage a larger number 
of turns or a larger flux is required than with a full-pitch winding. 
Fractional-pitch windings are used to a considerable extent both 
in direct- and in alternating-current machinery. 

Thus, the induced e.m.f . in an alternator or an induction motor 
is reduced by the distribution of the winding in more than one 
slot, and also by the use of a fractional-winding pitch. It is 
therefore convenient to consider the breadth factor kj, as being 
equal to the product of two factors, one taking into account the 
number of slots, and the other the influence of the winding pitch. 
We thus put 

K^kjc^, ....... (32) 

where ft, is called the slot factor and kw the winding-pitch factor. 

For a full-pitch winding kw= Ij and ki,=ka; for a fractional- 
pitch imislot winding kg=l, and kb=kw. The factors kg and kw 
are independent of one another, and their values are calculated 
in the next two articles. 

28. The Slot Factor k,. Let the stator of an alternator (or 
induction motor) have two slots per pole per phase, and let the 
centers of the adjacent slots be displaced by an angle a, in electri- 
cal degrees, the pole pitch, t, corresponding to 180 electrical 
degrees. If E (Fig. 17) is the vector of the effective voltage 
induced in the conductors in one slot, the voltage E^ due to the 
conductors in both slots is represented graphically as the geometric 
sum of two vectors E relatively displaced by the angle a. We see 
from the figure that \E' = E cos ^a, or E' = 2E cos ia. If both sets 
of conductors were bunched in the same slot we would then have 
E'= 2E. Hence, in this case the coefficient of reduction in 
voltage, or the slot factor, fc^=cosia. 

Let now the armature stamping have S slots per pole per phase, 
the angle between adjacent slots being again equal to a electrical 
degrees. Let the vectors marked E in Fig. 18 be the voltages 
induced in each slot; the resultant voltage E' is foimd as the geo- 



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Chap. IV] 



INDUCED E.M.F. 



69 



metric sum of the -B's. The radius of the circle r=iE/sinia, and 
iE' = r miiSa. Therefore, 

ks = E'/SE=i8m iSa)/S sin ia). . . . 



. . (33) 

When S=2j the formula (33) becomes identical with the expres- 
sion given before. 





Fig. 17. — ^A diagram illustrating the slot Fig. 18. — ^A diagram illustrating 
factor with two slots. the slot factor with several slots. 

The angle a depends upon the number of slots and the number 
of phases. Let there be m phases; then aSm= 180 degrees, and 

a = 1807(Sm) (34) 

The values of kg in.the table below are calculated by using the for- 
mulsB (33) and (34). 

VALUES OF THE SLOT FACTOR h 



Slots per Phase 
per Pole. 


Single-phase 
Winding. 


Two-phase 
Winding. 


Three-phase 


1 


1.000 


1.000 


1.000 


2 


0.707 


0.924 


0.966 


3 


0.667 


0.911 


0.960 


4 


0.653 


0.907 


0.958 


5 


0.647 


0.904 


0.957 


6 


0.643 


0.903 


0.956 


Infinity 


0.637 


0.900 


0.955 



In single-phase alternators part of the slots are often left empty 
so as to reduce the breadth of the winding and therefore increase 
the value of kg. For instance, if a punching is used with six slots 
per pole, perhaps only three or four adjacent slots are occupied. 
In this case, it would be wrong to take the values of kg from the 
first colunm of the table. If, for instance, three slots out of six 



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70 THE MAGNETIC CIRCUIT [Art. 29 

are occupied, the value of ka is the same as for a two-phase wind- 
ing with three slots per pole per phase. 

Prob. 20. Check some of the values of kg given in the table above. 

Prob. 21. The armature core of a single-phase alternator is built of 
stampings having three slots per pole; two slots per pole are utilized. 
What is the value of A;,? Ans. 0.866. 

Prob. 22. A single-phase machine has S uniformly distributed slots per 
pole, of which only /S' are used for the winding. What is the value of A;^? 
Ans. Use S' in eq. (33) instead of iS; preserve S in eq. (34). 

Prob. 23. A six-pole, 6600 v., Y-connected, 50-cycle turbo-alternator is 
to be built, using an armature with 90 slots. The estimated flux per pole 
is about 6 megalines. How many conductors are required per slot? 

Ans. 20. 

Prob. 24. What is the value of kg when the winding is distributed uni- 
formly on the surface of a smooth-body armature, each phase covering p 
electrical degrees? Solution : Referring to Fig. 18, kg is in this case equal 
to the ratio of the chord E' to the arc of the circle which it subtends. The 
central angle is ft and we have kg = (sinj/?) / (i/?7r/ 180°) . In a three-phase 
machine fi=QO degrees, and therefore A;, = 0.955. This is the value given 
in the last column of the table above. 

Prob. 25. Deduce the expression for kg given in the preceding problem 
directly from formula (33). Solution: Substituting Sa=P) S = ooand 
a =0, an indeterminate expression, O.oo, is obtained. But when the 
angle a approaches zero its sine is nearly equal to the arc, so that the 
denominator of the right-hand side of eq. (33) approaches the value 
iS.ia=iA where /9 is in radians. Changing fi to degrees, the required 
formula is obtained. 

29. The Winding-pitch Factor k^^ Let the distance between 
the two opposite sides of a coil (Fig. 16) be 180 —/^ degrees, where 
;- is the angle by which tlje winding-pitch is shortened. The volt- 
ages induced in the two sides of the coil are out of phase with 
each other by the angle ;-, so that if the voltage induced in each 
side is e, the total voltage is equal to 2e CosJ;' (Fig. 17). Fig. 17 
will apply to this case if we read y for the angle a. Hence, we 
have that 

ku,-=- cos i;- (35) 

In practice, the winding-pitch is measured in per cent, or as a frac- 
tion of the pole pitch r. For instance, if there are nine slots per 
pole and the coil lies in slots 1 and 8, the winding-pitch is 7/9, or 
77.8 per cent. If the coil were placed in slots 1 and 10 we would 
have a full-pitch, or a 100 per cent pitch winding. Let in gen- 



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Chap. IV] 



INDUCED E.M.F. 



71 



eral the winding-pitch be X,, expressed as a fraction. Then 
;'= (1 — ^)180°. Substituting this value of ;- into formula (35) we 
obtain 



*;«,= cos [90^(1 -g] 



(36) 



The values ot ky, given in Fig. 19 have been calculated according 
to this formula. 




0.70 



70 30 

Per Cent Winding Pitch 

Fig. 19. — Values of the winding-pitch factor. 



so 



100 



In applications, one takes the value of kg from the table, assum- 
ing the winding pitch to be one himdred per cent, and multiplies it 
by the value of k^ taken from the curve (Fig. 19). With frac- 
tional-pitch two-layer windings the value of kg corresponds to the 
number of slots per layer per pole per phase, and not to the total 
number of slots per pole per phase. This is clear from the explana- 
tion given in the preceding paragraph. Thus, for instance, in Fig. 
16, kg must be taken for three slots and not for five slots. If one 
has to calculate the values of k^ often, it is advisable to plot a set 



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72 THE MAGNETIC CIRCUIT [Art. 30 

of curves, like the one in Fig. 19, each curve giving the values of kb 
for a certain number of slots per pole per phase, against per cent 
winding pitch as abscissae. 

Prob. 26. In a 4-pole, 72-slot, turbo-alternator the coils lie in slots 1 
and 13. What is the per cent winding-pitch and by what percentage is 
the e.m.f. reduced by making the pitch short instead of 100%? 

Ans. 66.7 per cent; 1 — A^u; = 13.4 per cent. 

Prob. 27. What is the flux per pole at no load, in a 6600-volt, 25-cycle, 
500-r.p.m., Y-connected induction motor which has 90 slots, 36 conduc- 
tors per slot, and a winding-pitch of about 73 per cent? 

Ans. 7.26 megalines. 

Prob. 28. Show that for a chain winding kw is always equal to imity, 
in spite of the fact that some of the coils are narrower than the pole pitch. 

Prob. 29. Draw a sketch of a single-layer, fractional-pitch winding, 
using alternate slots for the overlapping phases. Show what values of kg 
and kto should be used for such a winding. 

30. Non-sinusoidal Voltages. In the foregoing calculations 
the supposition is made that the flux density in the air-gap is dis- 
tributed according to the sine law so that sinusoidal voltages are 
induced in each conductor. Under these circumstances the 
resultant voltage also follows the sine law, no matter what the 
winding-pitch and the number of slots are. The flux is practically 
sinusoidal in induction motors because the higher harmonics of 
the flux are wiped out by the secondary currents induced in the 
low-resistance rotor. But in synchronous alternators and motors 
with projecting poles the distribution of the flux in the air-gap is 
usually different from a pure sine wave. For instance, when the 
pole shoe is shaped by a cylindrical surface concentric with that of 
the armature, the air-gap length and consequently the flux density 
are constant over the larger portion of the pole; therefore, the 
curve of the field distribution is a flat one. This shape is improved 
to some extent by chamfering the pole-tips or by shaping the pole 
shoes to a circle of a smaller radius, so that the length of the air-gap 
increases gradually toward the pole-tips. 

When a machine revolves at a uniform speed, the e.m.f. induced 
in a single armature conductor has exactly the shape of the field- 
distribution curve, because in this case the rate of cutting the flux 
is proportional to the flux density (see eq. 27 above). There- 
fore, when a machine has but one slot per pole per phase (which 
condition is \mdesirable, but imavoidable in low-speed alternators, 
or in those designed for extremely high frequencies), the shape of 



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Chap. IV] INDUCED E.M.F. 73 

the pole-pieces must be worked out very carefully in order to have 
an e.m.f . approaching the true sine wave. With a larger number 
of slots this is not so necessary because the. em.fs* induced in differ- 
ent slots are added out of phase with each other, and the undesir- 
able higher harmonics partly cancel each other. The voltage 
wave is further improved by a judicious use of a fractional-pitch 
winding. These facts are made clearer in the solution of the prob- 
lems that follow.^ 

Prob. 30. The flux density in the air-gap under the poles of an alterna- 
tor is constant for 50 per cent of the pole pitch, and then it drops to zero, 
according to the straight-line law, on each side in a space of 15 per cent 
of the pole pitch. Draw to scale the curves of induced e.m.f. for the fol- 
lowing windings: (a) Single-phase, one slot per pole; (b) Single-phase, 
nine slots per pole, five slots being occupied by a one-hundred per cent 
pitch winding; (c) The same as in (b) only the winding-pitch is equal to 
7/9; (d) Three-phase, Y-connected full-pitch winding, two slots per pole 
per phase; in the latter case give curves of both the phase voltage and the 
line voltage. On all the curves indicate roughly the equivalent sine wave, 
in order to see the influence of the number of slots and of the fractional 
pitch in improving the wave form. 

Prob. 31. A three-phase, Y-connected alternator has three slots per 
pole per phase, and a full-pitch winding. The field curve has an 8 per 
cent fifth harmonic, that is to say, the amplitude of the fifth harmonic is 
0.08 of that of the fundamental sine wave. What is the magnitude of the 
fifth harmonic in the phase voltage and in the line voltage. Solution : In 
formula (33) the angle a between the adjacent slots is 20 electrical degrees 
for the fundamental wave. For the fifth harmonic the same distance 
between the slots corresponds to 100 electrical degrees. Hence, for the 
fundamental wave 

A;,=sin 30°/(3 sin 10°) -0.96; 

while for the fifth harmonic 

A;.,=sin 150°/(3 sin 50°) =0.217. 

This means that, due to the distribution in three slots, the fundamental 
wave of the voltage is reduced to 0.96 of its value in a unislot machine, 
while the fifth harmonic is reduced to only 0.217 of its corresponding value. 
Therefore, the relative magnitude of the fifth harmonic in the phase 
voltage is 8X21.7/96 = 1.8 per cent, which means that the fifth harmonic 
is reduced to less than one-fourth of its value in the field curve. In 
calculating the line voltage the vectors of the fundamental waves in a 
three-phase machine are combined at an angle of 120 degrees. Conse- 

^ For further details see Professor C. A. Adams' paper on " Electromotive 
Force Wave-shape in Alternators," Trans. Amer. Inst. Elec. Engs., Vol. 
28 (1909), Part II, p. 1053. 



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74 THE MAGNETIC CIRCUIT [Art. 31 

quently, the vectors of the fifth harmonic are combined at an angle 
of 120X5=600 degrees, or what is the same, — 120 degrees. Therefore 
the proportion of the fifth harmonic in the Une voltage is the same as 
that in the phase voltage. 

Prob. 32. Solve the foregoing problem when the winding pitch is 7/9. 
Ans. 0.33 per cent. This shows that by properly selecting the 
winding pitch an objectionable higher harmonic can be 
reduced to a negligible amomit. 

Prob. 33. Show that the line voltage of a Y-connected machine can 
have no 3d, 9th, 15th, etc. harmonics, that is to say, harmonics the num- 
bers of which are multiples of 3, no matter to what extent such harmonics 
are present in the induced e.m.fs. in each phase. 

Prob. 34. Prove that in order to have even harmonics in the induced 
e.m.f . of an alternator two conditions are necessary : (a) the flux distribu- 
tion under the alternate poles must be different; (b) the distribution of 
the armature conductors under the alternate poles must also be different 
from one another. Indicate pole shapes and an arrangement of the arma- 
tm^ winding particularly favorable for the production of the second har- 
monic. Noi€: In spite of a different distribution of flux densities the 
total flux is the same under all the poles. Therefore, the average voltages 
for both half cycles are equal (see Art. 24), though the shape of the two 
halves of the curve may be different, due to the presence of even har- 
monics. This shows that there is no " continuous- voltage component " 
in the wave, or rather that the voltage is in no sense unidirectional, and 
that a direct-current machine cannot be built with alternate poles without 
the use of some kind of a commutating device. 

31. The Induced E.M.F. in a Direct-current Machine. The 

e.m.f. induced in the armature coils of a direct-current machine 
(Fig. 20) is alternating, but due to the commutator, the voltage 
between the brushes of opposite polarity remains constant. 
This voltage is equal at any instant to the sum of the instantaneous 
e.m.fs. induced in the coils which are connected in series between 
the brushes. When a coil is transferred from one circuit to 
another, a new coil in the same electromagnetic position is intro- 
duced into the first circuit, and in this wise the voltage between 
the brushes is maintained practically constant, except for the small 
variations which occur while the armature is coming back to a 
symmetrical position. These variations are due to the coils short- 
circuited by the brushes and to the fact that the number of 
commutator segments is finite. 

Thus, to obtain the value of the voltage between the brushes, 
it is necessary to find the sum of the e.m.fs. induced at some 
instant in the individual armature coils which are connected in 



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Chap. IV] 



INDUCED E.M.F. 



75 



series between the brushes. Each e.m.f. represents an instanta- 
neous value of an alternating e.m.f.; the e.m.fs. induced in different 
coils differing in phase from one another, because they occupy 
different positions with respect to the poles. The voltages induced 
in the extreme coils of an armature circuit differ from one another 
by one-half of a cycle. 

Instead of adding the actual instantaneous voltages, it is suffi- 
cient to calculate the average voltage per coil, and to multiply it 
by the number of coils in series, because the wave form of the 
e.m.fs. induced in all the coils is the same, and their phase differ- 




FiG. 20. — ^The cross-section of a direct-current machine. 

ence is distributed uniformly over one-half of a cycle. According 
to eq. (26) the average voltage per turn per half a cycle is 2(^/^7", 
where iT is the time during which the coil moves by one pole 
pitch, and is the flux per pole, in webers. Substituting 1// for 
r, the average voltage per turn is equal to 4f0, Let there be N 
turns in series between the brushes of opposite polarity ; then the 
induced voltage of the machine is 



E=4fN0XlO-^, 



(37) 



where ^ is now in megalines. Thus, in a direct-current machine 
the induced voltage between the brushes depends only upon the 
total useful flux per pole, and not upon its distribution in the air- 
gap. 



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76 THE MAGNETIC CIRCUIT [Art. 31 

The relation between the number of turns in series and the total 
number of turns on the armature depends upon the kind of the 
armature winding.^ If the armature has a multiple winding N is 
equal to the total number of turns on the armature divided by the 
number of poles. For a two-circuit winding the number of turns 
in series is equal to one-half of the total number of turns. The num- 
ber of poles and the speed of the machine do not enter explicitly 
into formula (37), but are contained in the value of /. 

Prob. 36. A 90-8lot armature is to be used for a 6-pole, 580-r.p.m., 250- 
v., direct-current machine with a multiple winding. How many conduc- 
tors per slot are necessary if the permissible flux per pole is about 3 mega- 
lines? Ans. 10. 

Prob. 36. A 550-v., 4-pole railway motor has a two-circuit armature 
winding which consists of 59 coils, 8 turns per coil. The total resistance 
of the motor is 0.235 ohm. When the motor runs at 675 r.p.m. it takes in 
81 amp. What is the flux per pole at' this load? Ans. 2.5 ml. 

Prob. 37. Show that in a direct-current machine the use of a fractional- 
pitch winding has no effect whatever upon the value of the induced e.m.f ., 
as long as the winding-pitch somewhat exceeds the width of the pole shoe. 

Prob. 38. Prove that formula (37) is identical with the expression 

^ = (p/p')(r.p.m./60)C(^XlO-», (38) 

where C is the total number of armature conductors, p is the number of 
poles, and j/ is the number of circuits in parallel. 

Prob. 39. Show that the induced e.m.f. is the same when the armature 
conductors are placed in open or in closed slots as when they are on the 
surface of a smooth-body armature. See Prob. 18, Art. 26. 

Prob. 40. Considerable effort has been made to produce a direct-cur- 
rent generator with alternate poles, and without any commutator. One of 
the proposals which is sometimes urged by a beginner is to use an ordinary 
alternator, and to supply the exciting winding with an alternating current 
of the synchronous frequency. The apparent reasoning is that the field 
being reversed at the completion of one alternation the next half wave of 
the induced e.m.f . must be in the same direction as the preceding one, thus 
giving a miidirectional voltage. Show that such a machine in reality 
would give an ordinary alternating voltage of double the frequency. 
Hint : Make use of the fact that an alternating field can be replaced by two 
constant fields revolving in opposite directions. Or else give a rigid 
mathematical proof by considering the actual rate at which the armature 
conductors are cut by the field, which field is at the same time pulsating 
and revolving. 

* For details concerning the direct-current armature windings see the 
author's Experimental Electrical Engineering ^ Vol. 2, Chapter 30. 



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Chap. IV.] INDUCED E.M.F. 77 

Prob. 41. Prove that if in the preceding problem the frequency of the 
rotation of the poles is /i, and the frequency of the alternating current in 
the exciting winding is/a, that the voltage induced in the armature is a com- 
bination of two waves having frequencies of /i 4-/3 and /i —/a respectively. 

32. The Ratio of A.C. to D.C. Voltage in a Rotary Converter. 
A rotary converter resembles in its general construction a direct- 
current machine, except that the armature winding is connected 
not only to the commutator, but -also ±0 two or more slip rings,^ 
When such a machine is driven mechanically it can supply a direct 
current through its commutator, and at the same time an alter- 
nating current through its slip rings. It is then called a double- 
current generator. But if the same machine is connected to a 
source of alternating voltage and brought up to synchronous 
speed it runs as a synchronous motor and can supply direct ciurent 
through its commutator. It is then called a rotary converter. It 
is also sometimes used for converting direct current into alter- 
nating current, and is then called an inverted rotary. 

Both the direct and the alternating voltages are induced in the 
armature of a rotary converter by the same field, and oiu' problem 
is to find the ratio between the two voltages for a given arrange- 
ment of the slip rings. Consider first the simplest case of a single- 
phase converter with two collector rings connected to the arma- 
ture winding, at some two points 180 electrical degrees apart. If 
the armature has a multiple winding each slip ring is connected 
to the armature in as many places as there are pairs of poles. In 
the case of a two-circuit winding each collector ring is connected 
to the armature in one place only. 

If the machine has p poles then p times during each revolution 
the direct-current brushes make a connection with the same arma- 
ture conductors to which the slip rings are connected. At these 
moments the alternating voltage is a maximum, because the direct- 
current brushes are placed in the position where the induced volt- 
age in the armature is a maximum. Thus, with two slip rings, 
connected 180 electrical degrees apart, the maximum value of the 
alternating voltage is equal to the voltage on the direct-current 
side. If the pole shoes are shaped so that the alternating voltage 
^ approximately sinusoidal, the effective value of the voltage 
between the slip rings is 1/ V^=70.7 per cent of that between the 

* See the author's ExperirnenJUd Electrical Engineering j Vol. 2, Chapter 28. 



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78 THE MAGNETIC CIRCUIT [Art. 32 

direct-current brushes. The same ratio holds true for a two- 
phase rotary, for the voltages induced in each phase. 

Let now two slip rings be connected at two points of the arma- 
ture winding, a electrical degrees apart. In order to obtain the 

value of the alternating 
voltage the vectors of the 
^p voltages induced in the 

individual coils must be 
^^^;^>^ ^ \^ added geometrically, as in 

Fig. 18. With a large num- 
^1^ ^ ^ N ber of coils the chords can be 

Fenlaced bv the arc* and in 
Fig. 21. — ^Relation between the alternating \ •: . ^ . , 

voltages in a rotary converter. ^^^^ way Fig. 21 is obtained. 

The diameter Mti = e\ of the 
semicircle represents the vector of the alternating voltage when 
the points of connection to the slip rings are displaced by 180 
electrical degrees, while the chord AfP=e^ gives the voltage 
between two slip rings when the taps are distant by a electrical 
degrees. It will thus be seen that 

e^=6i sin ia (39) 

But we have seen before that 6i = 0.707 J5^, where E is the voltage 
on the direct-current side of the machine. Hence, for sinusoidal 
voltages, 

e„=0.707Esinia (40) 

The following table has been calculated, using this formula. 

Number of slip rings 2 3 4 5 6 

Angle between the adjacent taps in electrical 

degrees 180 120 90 72 60 

Ratio of alternating to continuous voltage, in 

percent 70.7 61.2 50 41.5 35.3 

The foregoing theory shows that the ratio of the continuous to 
the alternating voltage is fixed in a given converter, and in order 
to raise the value of the direct voltage it is necessary to raise the 
applied alternating voltage. This is done in practice either by 
means of various voltage regulators separate from the converter, 
or by means of a booster built as a part of the converter. Another 
method of varying the voltage is by using the so-called split-pole 
converter. In this machine the distribution of the flux density in 



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Chap. IV] INDUCED E.M.F. 79 

the air-gap can be varied within wide limits, and consequently 
that component of the field which is sinusoidal can be varied. The 
result is that the ratio of the direct to the alternating voltage is 
also variable. Namely, we have seen before that the value of the 
continuous voltage does not depend upon the distribution of the 
flux, but only upon its total value, while the effective value of the 
alternating voltage depends upon the sine wave or fimdamental 
component of the flux distribution.^ 

Prob. 42. Check some of the values given in the table above. 

Prob. 43. A three-phase rotary converter must deliver direct current 
at 550 V. What is the voltage on the alternating-current side? 

Ans. 337 v. 

Prob. 44. The same rotary is to be tapped in three additional places so 
as to get two-phase current also. How many different voltages are on 
the alternating-current side and what are they? 

Ans. 389, 337, 275, 195, 100. 

Prob. 45. The table given above holds true only when the flux density 
is distributed approximately according to the sine law. Show how to 
determine the ratio of alternating to continuous voltage in the case of two 
collector rings connected to taps 180 electrical degrees apart, when the 
curve of field distribution is given graphically. Solution : Divide the pole 
pitch into a sufficient number of equal parts and mark them on a strip of 
paper. Place the strip along the axis of abscissae. The sum of the ordi- 
nates of the flux-density curve, corresponding to the points of division, at 
a certain position of the strip, gives the instantaneous value of the alter- 
nating voltage. Having performed the siunmation for a sufficient number 
of positions of the strip, the wave of the induced e.m.f. is plotted. The 
scale of the curve is determined by the condition that the maximum ordi- 
nate is equal to the value of the continuous voltage. The effective value is 
foimd in the well-known way, either in rectangular or in polar coordinates 
(see the Electric Circuit). 

Prob. 46. Apply the solution of the preceding problem to the field dis- 
tribution specified in Prob. 30, Art. 30. Ans. 81.5 per cent. 

Prob. 47. Extend the method described in Prob. 45 to the case when 
the distance between the taps connected to the slip rings is less than 
180 degrees. Show how to find the scale of voltage. 

Prob. 48. How does a fractional pitch affect the values given in the 
table above, and the solution outlined in Prob. 45? 

Prob. 49. Show how to solve problems 45 to 48 when the field curve 
is given analytically, as B=F{a)y for instance in the form of a Fourier 
series. Hint: See C. A. Adams, " Voltage Ratio in Sjnichronous Conver- 
ters with Special Reference to the Split-pole Converter,'' Trans, Amer, 
Inst Elec, Engs,, Vol. 27 (1908), part II, p. 959. 

* See C. W. Stone, ** Some Developments in Synchronous Converters," 
Trans, Amer. Inst Elec, Engs., Vol. 27 (1908), p. 181. 



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CHAPTER V. 

THE EXCITING AMPERE-TURNS IN ELECTRICAL 
MACHINERY 

33. The Exciting Current in a Transformer. The magnetic 
flux in the core of a constant-potential transformer is determined 
essentially by the primary applied voltage, and is practically 
independent of the load (see Art. 25). When the terminal 
voltage is given, the flux becomes definite as well. The ampere- 
turns necessary for producing the flux are called the mag- 
netizing or the exciting ampere-turns. When the secondary cir- 
cuit is open the only current which flows through the primary 
winding is that necessary for producing the flux. This current is 
called the no4oad, exciting^ or magnetizing current of the trans- 
former. When the transformer is loaded, the vector difference 
between the primary and the secondary ampere-turns is practi- 
cally equal to the exciting ampere-turns at no-load. 

The exciting current is partly reactive, being due to the 
periodic transfer of energy between the electric and the magnetic 
circuits (see Art. 16 above), partly it represents a loss of energy 
due to hysteresis and eddy currents in the core. Sonie writers call 
the reactive component of the no-load current the magnetizing 
current, and the total no-load current the exciting current. 
Generally, however, the words magnetizing and exciting are used 
interchangeably to denote the total no-load current. The 
components of the current in phase and in quadrature with the 
induced voltage are called the energy and the reactive components 
respectively. 

The no-load or exciting current in a transformer must usually 
not exceed a specified percentage of the rated full-load current; 
it is therefore of importance to know how to calculate the 
exciting current from the given dimensions of a transformer. 
Knowing the applied voltage and the number of turns, the maxi- 
mum value of the flux is calculated from eq. (28). We shall 

80 



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Chap. V] EXCITING AMPERE-TURNS 81 

assume first that the maximum flux density in the core is so 
low, that it lies practically on the straight part of the mag- 
netization curve of the material (Fig. 3). The case of high flux 
densities is considered in the next article. 

Since by assumption the instantaneous magnetomotive forces 
are proportional to the corresponding flux densities, the magnetiz- 
ing current must vary according to the sine law. It is suflicient, 
therefore, to calculate the maximum value of the magnetomotive 
force, corresponding to the maximum flux. Knowing the ampli- 
tude ^m of the flux and the net cross-section of the core, A, the 
flux density B^ becomes known; from the magnetization curve of 
the material (Fig. 3) the corresponding value of Hm, or the ampere- 
turns per imit length of path, is foimd. The mean length I of the 
lines of force is determined from the drawing of the core, so that 
the total magnetizing ampere-turns Mm=Hml can be calculated. > 
The mean magnetic path aroimd the comers is somewhat shorter 
than the mean geometric path. 

Let ni be the number of turns in the primary winding, and io 
the effective value of the reactive component of the exciting cur- 
rent. We have then 

ioni\/2-=Mm. . .• (41) 

From this equation the quantity which is unknown can be calcu- 
lated. 

It is presupposed in the above deduction that the joints 
between the laminations offer no reluctance. In reality, 
the contact reluctance is appreciable; its value depends upon 
the character of the joints, and the care exercised in the 
assembling of the core. This reluctance of the joints can be 
expressed by the length of an equivalent air-gap having the same 
reluctance. Thus, experiments show that each overlapping joint 
is equivalent to an air-gap 0.04 mm. long. A butt joint, with 
very careful workmanship, is equivalent to an air-gap of about 
0.05 mm.; in practice, a butt joint may offer a reluctance of from 
50 to 100 per cent higher than the foregoing value.^ Knowing the 

* H. Bohle, "Magnetic Reluctance of Joints in Transforming Iron," Journal 
(British) Inst. Electr, Engs., Vol. 41, 1908, p. 627. It is convenient to 
estimate the influence of the joints in ampere-tums at a standard flux 
density. For each lap joint 32 ampere-turns must be added at a density 
of 10 kilolines per square centimeter, while a butt joint requires at the same 



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82 THE MAGNETIC CIRCUIT [Art. 33 

length oof the equivalent air-gap, the number of additional ampere- 
turns is calculated according to the formula aBm/ ti, and then is 
multiplied by the number of joints in series (usually four). This 
number of ampere-turns must be added to Af^ calculated above. 
The energy component ii of the exciting current is determined 
from the power lost in hysteresis and eddy currents in the core. 
Having calculated this power P as is explained in Article 19, we 
find i\ = P/Ei, where E\ is the primary applied voltage. Knowing 
io and ii, the total no-load current is foimd as their geometric sum, 

The watts expended in core loss depend only upon the volume 
of the iron, the frequency, and the flux density used. It can be 
also shown that the reactive volt-amperes required for the excita- 
tion of the magnetic circuit of a transformer depend only upon the 
volume of the iron, the frequency, and the flux density. Namely, 
neglecting the influence of the joints, eq. (41) can be written in the 
form 

toni\/2=HJ. 

Eq. (28) in Art. 25 can be written as 

£i=4.44ni/AB^XlO-5 

where A is the cross-section of the iron, and B^ is the maximum 
flux density, in kilolines per square centimeter. Multiplying these 
two equations together, term by term, and cancelling n\ we get, 
after reduction, 

Siio/^=^/SmH«XlO-5, .... (42) 

where V=Al is the volume of the iron, in cubic centimeters. 
The left-hand side of eq. (42) represents the reactive magnetizing 
volt-amperes per imit volume of iron; the right-hand side is a 
fimction of / and B^, only, because Hm, can be expressed through 
Bm from the magnetization curve of the material. 

Formula (42) can be plotted as a set of curves, one for each 
commercial frequency. These curves are quite convenient in the 
design of transformers, because they enable one to estimate 
directly either the permissible volume of iron, or the permissible 
fl\ix density, when the reactive component of the exciting cur- 
density from 60 to 80 ampere-tums. At other flux densities the increase 
is proportional. 



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Chap. V] EXCITING AMPERE-TURNS 83 

rent ,is limited to a certain percentage of the full-load current. In 
practice, such curves are sometimes plotted directly from the 
results of tests on previously built transformers. These experi- 
mental curves are the most secure guide for predicting the exciting 
current in transformers; formula (42) shows their rational basis. 

Prob. 1. Prove that if there were no core loss the exciting current 
would be purely reactive, that is to say, in a lagging phase quadrature 
with the induced voltage. 

Prob. 2. The core of a 22-kv. 25-cycle transformer, like the one 
shown in Fig. 12, has a gross cross-section of 4500 sq.cm.; the mean 
path of the lines of force is 420 cm.; the material is silicon steel; the 
maximum flux is 36 megalines. The expected reluctance of each of the 
four butt joints is estimated to be equivalent to an 0.08 mm. air-gap. 
What are the two components of the exciting current, and what is the 
total no-load current? Ans. 1.8; 0.4; 1.85. 

Prob. 3. In what respects does the calculation of the magnetizing 
current in a shell-type or cruciform-type transformer differ from that 
in a core-type transformer? 

Prob. 4. Show that for flux densities up to 10 kl. /sq.cm. the mag- 
netizing volt-amperes per kilogram of carbon steel at 60 cycles are 
approximately equal to 7.3(Bm/10)^ 

Prob. 6. Show that the influence of the joints can be taken into 
account in formula (42) by adding to the actual volume of the iron 
the volume of the air-gaps multiplied by the relative permeability of 
the iron. 

Prob. 6, A shell-type 1000-kva., 60-cycle transformer is to have a 
core made of silicon-steel punchings of a width w = 17 cm. (Fig. 13); 
the average length of the magnetic path in iron is 180 cm. ; the reactive 
component of the no-load current must not exceed 2 per cent of the 
full-load current. Draw curves of the required height of the core per 
link, and of the total core loss in per cent of the rated kva., for flux 
densities up to 10 kl./sq.cm. 

Ans. B'/i-5200; at 5=9, P=0.51 per cent. 

34. The Exciting Current in a Transformer with a Saturated 
Core. In the preceding article the flux density in the core is 
supposed to be within the range of the straight part of the satura- 
tion curves (Fig. 3), so that, when the flux varies according to the 
sine law, the magnetizing current also follows a sine wave. We 
shall now consider the case when the flux density rises to a value 
on or beyond the knee of the magnetization curve. Such high flux 
densities are used with silicon steel cores, especially at low frequen- 
cies. In this case the magnetizing current does not vary according 
to the sine law, but is a peaked wave, because at the moments when 



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84 



THE MAGNETIC CIRCUIT 



[Art. 34 



the flux is approaching its maximum, the current is increasing 
faster than the flux, on account of saturation. The amplitude 
factor of the current wave, or the^ ratio of the amplitude to the 
effective value is no more equal to V2, but is larger. Let this 
ratio be denotedby Xa- Then eq. (41) becomes 



ioniXa='M„ 



(43) 



where iq is as before the effective value of the reactive component 
of the exciting current. The value of Xa is obtained by actually 



3.0 



2.5 



u2.0 



3U 



1.0 



0.5 



ao 
























-^ 


jg^^""^--^" """"*'■- "■" 


■vC^ 


^-«^^ "-^ 


^^^ 


p< P ^ ' 


) X-W 


'^ 


^t^ 




^^^ 




^^^ 




^^^^ 








— — _ — — ' '^ 




























































6 10 li 


S 20 26 



Plux density in KL per Sq. Cm. 

Fig. 22. — ^Ratio of the amplitude to the effective value of the magnetizing 

current. 

plotting the curve of the magnetizing current from point to point 
and calculating its effective value. Since the procedure is rather 
long, it is convenient to calculate the values of Xa once for all for 
the working range of values Bm* This has been done for the mate- 
rials represented in Fig. 3, and the results are plotted in Fig. 22. 

Strictly speaking, the exciting current is imsymmetrical, due 
to the effect of hysteresis, and the values of Xa ought to be cal- 
culated, using the hysteresis loops of the steel. However, it is very 
nearly correct to calculate Xa from the magnetization curve, and to 



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Chap. V] EXCITING AMPERE-TURNS 86 

calculate the energy component of the exciting current separately, 
from the core loss curves (Fig. 10). The magnetizing current 
required for the joints is calculated separately, using ;ra=v'2, 
according to eq. (41). The total effective magnetizing current is 
foimd by adding together the values of io for the iron and for the 
joints. The loss component, ii, is added to this value in quadra- 
ture, to get the total no-load current. As is mentioned above, it 
is preferred in practice to estimate the total exciting current of new 
transformers from the curves of no-load volt-amperes per kilogram 
of iron, the values being obtained from tests on similar trans- 
formers. 

Prob. 7. The core of a 25-cycle cruciform type transformer (Fig. 14) 
weighs 265 kg.; the mean length of the magnetic path is 170 cm.; the 
material is silicon steel. The 4400-v. winding of the transformer has 
1100 turns in series. What is the reactive component of the no-load 
current? Ans. 8.4 amperes. 

Prob. 8. Check a few points on the curves in Fig. 22. 

Prob. 9. Show that in formula (42) the coefficient tt is a special 
case of the more general factor 4.44/;|;a, when the magnetizing current 
does not follow the sine wave. 

Prob. 10. What are the reactive volt-amperes per kilogram of carbon 
steel at 40 cycles and at a flux density of 16 kl./sq.cm.? Ans. 56.4. 

Prob. 11. Show how to calculate the exciting ampere-tums required 
for a given flux in a thick and short core in which the flux density is 
different along different paths. 

35. The Types of Magnetic Circuit Occurring in Revolving 
Machinery. The remainder of this chapter and the next chapter 
have for their object the calculation of the exciting ampere-tums 
necessary for producing a certain useful flux in the principal types 
of electric generators and motors. In direct-current machines, 
in alternators, and in rotary convertors it is necessary to know the 
exciting or field ampere-tums in order to plot the no-load satura- 
tion curve, to predict the performance of the machine imder vari- 
ous loads, and to design the field coils. In an induction motor one 
wants to know the required excitation in order to determine the 
no-load current, or to calculate the number of turns in the stator 
winding, when the limiting value of the no-load current is pre- 
scribed. The general procedure in determining the required 
number of ampere-tums for a given flux is in many respects the 
same in all the types of electrical machinery, so that it is possible 
to outline the general method before going into details. 



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86 



THE MAGNETIC CIRCUIT 



[Art. 35 



In direct-current machines and in synchronous generators, 
motors, and rotary converters, the magnetic flux (Figs. 15 and 20) 
from a field pole passes into the air-gap and the armature teeth. 
In the armature core the flux is divided into two halves, each half 
going to one of the adjacent poles. The magnetic paths are com- 
pleted through the field frame. Part of the flux passes directly 
from one pole to the two adjacent poles through the air, without 
going through the armature. This part of the flux is known as 
the leakage flux. The closed magnetic paths and the field coils of a 
machine may be thought of as the consecutive links of a closed 
chain. While in a transformer the chain is open, in generators 




Fig. 23. — The paths of the main flux and of the leakage fluxes in an 
induction motor (or generator). 

and motors the chain must be closed on accoimt of the continuous 
rotation. 

In induction machines, both generators and motors (Fig. 23), 
the flux at no load is produced by the currents in the stator wind- 
ings only. When the machine is loaded, the flux is produced by 
the combined action of the stator and rotor currents, the rotor cur- 
rents opposing those in the stator, the same as in a transformer. 
Therefore, the flux in the loaded machine may be regarded as the 
resultant of the following three component fluxes: The main or 
useful flux, <P, which links with both the primary and the secondary 
windings; the primary leakage flux, <Pi, which links with the stator 
winding only; and the secondary leakage flux, 02 which is linked 
with the rotor winding alone. The leakage fluxes not only do not 



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Chap. V] EXCITING AMPERE-TURNS 87 

contribute to the useful torque of the machine, but actually reduce 
it. In reality, there is of course but one flux, the resultant of the 
three, but for the purposes of theory and computations the three 
component fluxes can be considered as if they had a real separate 
existence. In this and in the following chapter the main flux 
only will be discussed for this type of machinery. Considera- 
tion of the leakage flux will be reserved to Art. 66. 

The total magnetomotive force per magnetic circuit is equal 
to the sum of the m.m.fs. necessary for estabUshing the required 
flux in the separate parts of the circuit which are in series, viz., the 
pole-pieces, the air-gap, the teeth, and the armature core. All the 
necessary elements for the solution of this problem have been dis- 
cussed in the first two chapters. It remains here to establish some 
semi-empirical " short-cut " rules and formulae for the irregular 
parts of the circuit, for which, although close approximations can 
be made, the exact solution is either impossible or too complicated 
for the purposes of this text. The following topics are considered 
more in detail in the subsequent articles of this and of the follow- 
ing chapter. 

(a) The ampere-turns necessary for the air-gap when it is 
limited on one side or on both sides by teeth, so that the flux den- 
sity in the air-gap is not imiform. 

(6) The ampere-turns necessary for the armature teeth when 
they are so highly saturated that an appreciable part of the flux 
passes through the slots between the teeth. 

(c) The ampere-turns necessary for the highly saturated cores 
in which the lengt>hs of the individual paths differ considerably 
from one another, with a consequent lack of imiformity in the flux 
density. 

(d) The leakage coefficient and the value of the leakage flux 
which passes directly from pole to pole. This leakage flux 
increases the flux density in the poles and in the field frame of the 
machine, and consequently increases the required number of 
ampere-turns. 

All of the m.m.f. calculations that follow are per pole of the 
machine, or what is the same, for one-half of a complete magnetic 
circuit {cdfg in Figs. 15, 20, and 23), the two halves being identical. 
This fact must be borne in mind when comparing the formulae with 
those given in other books, in which the required ampere-turns are 
sometimes calculated for a complete magnetic circuit. 



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THE MAGNETIC CIRCUIT 



[Art. 36 



Prob. 12. Inspect working drawings of electrical machines found 
in various books and magazine articles; indicate the paths of the main 
and of the leakage fluxes; and make clear to yourself the reasons for the 
use of different kinds of steel and iron in the frame, the core, the pole- 
pieces, and the pole shoes. 

Prob. 13. Make sketches of the magnetic circuit of a turbo-alternator 
with a distributed field winding, of a homopolar machine, of an inductor- 
type alternator, and of a single-phase commutator motor. Indicate the 
paths of the useful and of the leakage fluxes. 

36. The Air-gap Ampere Turns. The general character of the 
distribution of the magnetic flux in the air-gap of a synchronous 
and of a direct-current machine is shown in Fig. 24, the curvature 
of the armature being disregarded. The principal features of this 
flux distribution are as follows: 



Armature 



^Air-duct 



-mature 




Fig. 24. — ^The cross-section of a direct-current or synchronous machine, 
showing the flux in the air-gap. 

(a) The flux per tooth pitch A is practically the same under all 
the teeth in the middle part of the pole, where the air-gap has a 
constant length, and is smaller for the teeth near the pole-tips 
where the air-gap is larger. 

(b) On the armature surface the flux is concentrated mainly 
at the tooth-tips; very few lines of force enter the armature 
through the sides and the bottom of the slots. 

(c) There is a considerable spreading, or fringing, of the lines 
of force at the pole-tips. 

(d) In the planes passing through the axis of the shaft of the 
machine there is also some spreading or fringing of the lines of 
force at the flank surfaces of the armature and the pole, and in 
the ventilating ducts. 



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Chap. V] EXCITING AMPERE-TURNS 89 

This picture of the flux distribution follows directly from the 
fundamental law of the magnetic circuit, the flux density being 
higher at the places where the permeance of the path is higher. 
The actual flux distribution is such that the total permeance of all 
the paths is a maximum, as compared to any other possible distri- 
bution. In other other words, the flux distributes itself in such a 
way, that with a given m.m.f . the total flux is a maximum, or with 
a given flux the required m.m.f. is a minimum. This is confirmed 
by the beautiful experiments of Professor Hele-Shaw and his col- 
laborators,^ who have obtained photographs of the stream lines of a 
fluid flowing through an arrangement which imitated the shape 
and the relative permeances of the air-gap and of the teeth in an 
electric machine. 

Let (Pa be the total permeance in perms of the air-gap between 
the surface of the pole shoe and the teeth, and let be the useful 
flux per pole, in maxwells, which is supposed to be given. Then, 
according to eq. (2), Art. 5, the number of ampere-turns required 
for the air-gap is 

Ma=(P/(Pa (44) 

The problem is to calculate the permeance of the gap from the 
drawing of the machine. 

One of the usual practical methods is to calculate (P imder cer- 
tain simplifying assumptions and then multiply the result by an 
empirical coefficient determined from tests on similar machines. 
The simplest assumptions are (Fig. 25) : (a) that the armature has 
a smooth surface, the slots being filled with iron of the same per- 
meability as that of the teeth; (6) that the external surface of the 
pole shoes is concentric with that of the armature; (c) that the 
equivalent air-gap aeq is equal to two-thirds of the minimum air- 
gap plus one-third of the maximum air-gap of the actual machine; 
(d) that the ventilating ducts are filled with iron; (e) that the paths 
of the fringing flux at the edges of the pole shoe are straight lines, 
and extend longitudinally to the edge of the armature surface and 
laterally for a distance equal to the equivalent air-gap on each 
side. 

^ For a detailed account of the experimental and theoretical investigations 
on this subject, with numerous references, see Hawkins and Wallis, The 
Dynamo (1909), Vol. 1, Chapter XV. 



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90 



THE MAGNETIC CIRCUIT 



[Art. 36 



With these assumptions, the permeance of the " simplified " 
air-gap is 

(P.= 1.25w;.I./o#«, (45) 

where Wg is the average width of the flux, and I, is its average 
axial length. Or 

W9=i(Wa+Wp)=Wp+a'; 

The lateral spread a' of the lines of force at each pole-tip is 
taken to be approximately equal to Oe^. 



^ 



-Wtr- 






-hr 



T 



"^ 



llllllllllll 



r 



Fig. 25. — ^Magnetic flux in the simplified air-gap. 

The permeance of the actual air-gap is smaller than that of the 
simplified gap, so that we have 

^a^^Aa, (46) 

where k^ is a coefficient larger than unity, called the air^ap factor. 
Substituting the value of (P from eq. (46) igito (44) gives 



M^=h 



(47) 



so that ka is the factor by which the ampere-turas for the simplified 
air-gap must be multiplied in order to obtain the ampere-tums 
required for the actual air-gap. The value of ka usually varies 
between 1.1 and 1.3, depending on the relative proportions of the 
teeth, the slots, and the air-gap, and on the shape of the poles. 

The numerical values of fcaare calculated from the results of tests 
on machines of proportions similar to that being computed. Let 
the no-load saturation curve of a machine be available from test; 
this is a curve which gives the relation between the induced voltage 



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Chap. V] EXCITING AMPERE-TURNS 91 

and the field current of the machine. From the known specifica- 
tions of the machine this curve can be easily converted into one 
which gives the useful flux per pole against the ampere-turns per 
pole as abscissae. The lower part of such a curve is always a 
straight line, there being then practically no saturation in the iron. 
On this part of the curve, practically the whole m.m.f . is consumed 
in the air-gap, so that the actual permeance of the air-gap is found 
by dividing one of the ordinates by the corresponding abscissa. 
The permeance of the simplified air-gap is calculated from eq. (45), 
and the ratio of the two gives the value of the coefficient fco. This 
value is then used in the design and calculation of the performance 
of new machines with similar proportions. Engineering judg- 
ment and practical experience are factors of considerable impor- 
tance in estimating the values of ka for new machines. 

The same method of calculating the air-gap ampere-turns is 
applicable to induction machines (Fig. 23). The ampere-turns 
are computed, assuming both the rotor and the stator to have 
smooth iron surfaces, without slots; the result is then multiplied 
by a factor ka larger than imity, determined from tests upon 
machines of similar proportions. 

A more accurate, though more elaborate, method for calcula- 
ting the air-gap ampere-turns is explained in the next article. 

Prob. 14. Calculate the air-gap ampere-turns per pole for a 6600 v., 
25-cycle, 375-r.p.m. alternator to be built according to the following 
specifications: The bore 2.4 m.; the gross axial length of the armature 
core 55 cm.; seven air ducts 9 mm. each; the minimum air-gap is 15 
mm. ; the maximum air-gap is 30 mm. The poles cover 66 per cent of 
the periphery; the axial length of the pole shoes is 53 cm. The useful 
flux per pole at no-load and at the rated voltage is 19.1 megalines. The 
air-gap factor is estimated to be about 1.15. Ans. 10,600. 

Prob. 16. The no-load characteristic obtained from the test upon 
the machine specified in the preceding problem has a straight part such 
that at a field current of 52 amp. the line voltage is 4000 v. Each field 
coil has 120 'turns. What is the true value of the air-gap factor? 

Ans. 1.12. 

Prob. 16. A pole shoe is so shaped that the minimum air-gap is Oq 
and the maximum air-gap is Oj^Oo+Jo, the increase in the length being 
proportional to the square of the distance from the center of the pole. 
What is the length a^q of the equivalent uniform air-gap such that its 
total permeance is the same as that of the given air-gap? Assume a 
smooth-body armature, and neglect the fringing at the pole-tips. Solu- 
tion: Let the peripheral width of the pole be 2w; then the length of 



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92 THE MAGNETIC CIRCUIT [Akt. 37 

the air-gap at a distance x from the center is a =«© ■^^O'ix/w)'^. The per- 
meance of an infinitesimal path of the width dx is proportional to dx/ax. 
Hence we have the relation 

Jdx/[an ■\-Aa{x/w) ']= w/aeq, 


from which 

a^q = Voo^o/tan"-^ (Vjo/oo) . 

Prob. 17. What is the length of the equivalent air-gap in the preced- 
ing problem if the clearance at the pole- tips is twice the clearance at 
the center of the pole? Ans. 1.273ao. 

Prob. 18. Show that, when the air-gap is non-uniform, the length 
of the equivalent uniform gap can be determined approximately, accord- 
ing to Simpson's Rule, from the equation 

where a©, ai, and Om are the lengths of the gap at the center, at the tip of 
the pole, and midway respectively. If the air-gap is imiform under the 
major portion of the pole, but the pole shoe is chamfered, more terms 
must be taken in Simpson's formula in order to obtain a^q with a 
sufficient accuracy. 

Prob. 19. What is the length of the air-gap required in problems 
16 and 17, according to the formula given in problem 18? 

Ans. 1.276ao. 

37. The Method of Equivalent Permeances for the Calculation 
of Air-gap Ampere-turns. An inspection of Fig. 24 will show that 
the total permeance of the air-gap is made up of a number of per- 
meances in parallel. It is equal therefore to the sum of these 
permeances. For the purpose of calculation two kinds of per- 
meances are considered separately: those from the teeth to 
the pole surface proper, and those from the teeth to the pole-tips. 
The former can be calculated quite accurately, the latter are to 
some extent estimated. 

The permeance per tooth pitch in the part of the air-gap near 
the center of the pole can be divided into two parts, that imder the 
tooth-tip, and the fringe from the sides of the slots and in the ven- 
tilating ducts. The permeance of the paths which proceed from 
the tooth-tip constitutes the larger portion and is made up of 
nearly parallel lines; this permeance is therefore easily computed. 
The values of the permeance of the fringe from the flank of the 
tooth to the perpendicular surface of the pole have been deter- 



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Chap. V] EXCITING AMPERE-TURNS 93 

mined theoretically by Mr. F. W. Garter.^ Only the numerical 
resiilts are given here, in a somewhat simplified practical form; 
the solution itself presupposing a knowledge of the properties of 
conjugate fimctions 2 

Consider the permeance of two tooth fringes, such as o^pgr and 
Qfpfq^'ff (Fig^ 24), perpendicular to the plane of the paper. This 
permeance depends only upon the ratio of the slot width s to the 
length a of the air-gap, for let both s and a be increased say twice: 
The length and the cross-section of each elementary tube of force 
is also increased twice, hence its permeance remains the same. 

The permeance of each fringe can be replaced by the permeance 
of an equivalent rectangular path of the length a and of a width 
\At (Fig. 26). This is the same as increasing the width of the 
tooth by the amount At and assuming all the lines of force to be 
parallel to each other in the air-gap. The permeance of the path 
which replaces the two fringes is equal to iiAt/a. From what has 
been said^above follows that the ratio At/ a depends only upon the 
ratio of s/a; the relationship between the two ratios is plotted in 
Fig. 26, from Carter's calculations. For the sake of convenience 
and accuracy, the curve is drawn to two different scales, one for 
large the other for small values of s/o. 

The curve in Fig. 26 may be interpreted in two ways: It may 
be said to represent the ''geometric permeance" of the fringe (for 
11= \)] or else it may be said to give the correspondings sets of 
values of s and Atj measured in the lengths of the air-gap as the 
unit. With a given a, At increases with s, because the maximum 
width of the actual fringe is is. With a given s the width At 
increases toward the pole-tip (if the air-gap is variable), because 
with a longer air-gap the fringing lines of flux fill a larger part of 
the air-gap imder the slot. 

The corrected width of the tooth hV=t'\-At and the permeance 
of the air-gap, in perms per tooth pitch, is 

(Pat=l^2b{At/a^^-t/a^)l^y .... (48) 

*Note on Air-gap Induction, Joum, Inst. Electr. Eng. (British), Vol. 29, 
(1899-1900), p. 929; Air-gap Induction, Electrical Worldy Vol. 38, (1901) 
p. 884; See also Hawkins and Wallis, The Z>ynamo (1909), Vol. 1, p. 446; 
E. Arnold, Die Gleichstrommaschine (1906), Vol. 1, p. 266. 

^ J. C. Maxwell, Electricity and Magnetism, Vol. 1, p. 284; J. J. Thomson, 
Recent Researches in Electricity and Magnetism f Chapter III; Horace Lamb, 
Hydrodynamics (1895), Chapter IV. 



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94 THE MAGNETIC CIRCUIT [Art. 37 

where a^ is the length of the air-gap at the center of the tooth, and 
leif is the effective axial length of the machine (see below). The 
value of At/ ax must be taken from Fig. 26 for the corresponding 
ratio s/Oj.. 

The permeance of the pole fringe, hmn in Fig. 24, cannot be cal- 
culated by the foregoing method, because this permeance depends 
upon the irregular shape of the pole-tip. The pole-fringe permeance 
is usually estimated graphically by drawing lines of force, taking 
Fig. 24 as a guide^ ; the permeance of each tube of flux between the 
pole and the armature is ;l4//, where A is the mean cross-section, 
and I is the mean length of the tube. The fringe permeance is of 
the order of magnitude of 10 per cent of the total* permeance of 
the air-gap, so that some error in its estimation does not seriously 
affect the total required ampere-turns. Careful designers some- 
times calculate the air-gap permeance for two positions of the 
pole, differing from each other by one-half of the tooth pitch, and 
take the average of the two results. 

Carter's curve could be used directly for calculating the pole- 
fringe permeance, if the pole waist were of the same width as 
the pole shoe (line mmf in Fig. 24), and if the armature had no 
slots. In this case the space between the adjacent poles could be 
considered as a big slot, and the curve in Fig. 26 could be directly 
applied to it. On account of a smaller width of the pole core and 
because of the armature slots the mean length of the lines of force 
in the fringe is increased, so that the actual permeance of the pole- 
fringe is somewhat smaller than that according to Carter's curve. 
By practice and experience one can acquire a judgment as to what 
fraction of Carter's permeance to take in a given case. 

The length leff is a sum of the parts such as Zi, h, etc. (Fig. 24), 
on which the lines of force are parallel, and of small additional 
lengths which take accoimt of the fringing in the air-ducts and at 
the pole flanks. These additional lengths are again estimated 
from Carter's curve (Fig. 26). The fringe Ad in an air-duct of 
the width d is practically the same as that in a slot of the width 
s=d. The additional length J/ for the pole flanks is found by con- 
sidering the two fringes as due to a slot of the width /. When the 
stationary and the revolving parts are of the same axial length so 
that /=0, there still remains some fringe permeance between the 

* See Art. 41 below in regard to the drawing of the lines of force by the 
judgment of the eye. 



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Chap. Vl 



EXCITING AMPERE-TURNS 



96 





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96 THE MAGNETIC CIRCUIT [Art. 37 

flank surfaces of the two iron structures. This permeance is, how- 
ever, very small, and has to be estimated empirically, if at all. 

Strictly speaking, leff is different for each tooth, if the air-gap 
is variable, because the amoimt of fringing in the air-duqts and at 
the flanks is different. However, it is hardly worth the effort 
in ordinary cases to calculate left for each tooth. It is sufficient to 
take an average Z«// for some intermediate value of the air-gap. 

In some high-speed alternators, and usually in induction 
motors, air-ducts are provided in both the stationary and the 
revolving parts, in the same planes. The flux fringe in an air-duct 
is then of such a shape that the lines of force are parallel to one 
another in the middle of the air-gap, between the stator and the 
rotor. Therefore, when using the curve in Fig. 26 for such a case, 
the cylindrical surface midway between the- stator and the rotor 
must be taken to correspond to that of the solid iron surface 
assumed in the deduction of the curve. Hence, iax must be used 
instead of a^ in determining JL 

Having calculated the permeances of the several paths per pole 
pitch the total permeance of the air-gap is f oimd as their sum, or 

'(Pa=^(Pat+2(Pfrinae (49) 

Then, the required number of ampere-turns is determined from 
eq. (44). The method gives quite correct results, especially with 
some experience in estimating the permeances of irregular paths. 
Each designer usually modifies slightly the empirical factors which 
are indispensable in this method, and devises short cuts good for 
the particular kind of machine in which he is interested. 

Instead of calculating the permeance of each tooth separately, 
some engineers replace the actual variable air-gap by an equiva- 
lent constant air-gap aeg, either by the judgment of the eye, or as 
in prob. 18 above. The actual peripheral length of the pole arc is 
increased by from one to one and one-half aeq on each side to take 
into accoimt the fringing at the pole-tips. This gives the number 
of teeth imder the pole. The permeance of each tooth is calcu- 
lated from eq. (48) for ax = aeqy and is then multiplied by the num- 
ber of teeth. With some practice, one can obtain in this manner 
quite accurate results at a considerable saving in time. 

The method outlined above is not directly applicable to induc- 
tion machines which have slotted cores on both sides of the air- 



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Chap. V] EXCITING AMPEKE-TURNS 97 

gap (Fig. 23) : At each instant some stator teeth are opposite rotor 
teeth, others bridge over some rotor slots, and vice versa. The 
amoimt of overlap varies from instant to instant, causing periodic 
fluctuations in the air-gap reluctance. 

Assume first that both the stator and the rotor have smooth 
surfaces facing the air-gap. Let the permeance of such a machine 
be (P«. If now the armature be slotted, the cross-section of 
the paths in the air-gap (neglecting the fringe) is reduced in the 
ratio ti/Xi where ti and Xi are the stator tooth width and tooth 
pitch respectively. The permeance (P^ is also reduced in the same 
ratio. Let the rotor be also provided with slots; the average 
cross-section of the path is thereby further reduced in the ratio 
(^2/^2), where ^2 and ^2 are the tooth width and the tooth pitch on 
the surface of the rotor. Thus, disregarding the spread of the 
flux, the average air-gap permeance of an induction motor is 

(^a) = «lA)(feA)^a, 

the symbol (Pa being put in parentheses to indicate that a further 
correction for the tooth fringe is necessary.^ 

In order to take the fringe into consideration, an empirical cor- 
rection is made in this formula. Namely, it is assumed that the 
actual permeance of the fringes of the stator teeth is the same as 
if the rotor had a smooth core, and vice versa. Accordingly, in the 
preceding formula, the values ti and <2 of the tooth widths are 
corrected for the fringe, using Carter's curve (Fig. 26). The 
formula becomes then 

^a=(<l'A)(«27A2)tP., (50) 

where i/ and fe' are the corrected widths of the stator and rotor 
teeth respectively. This formula has been found to be in a satis- 
factory agreement with experimental results.^ 

* For a more rigorous proof of this formula see C. A. Adams, "A Study 
in the Design of Induction Motors," Trans. Amer. Inst. EUctr. Engs., Vol. 24 
(1905), p. 335. 

2 T. F. Wall, The Reluctance of the Air-gap in Dynamo-machines, Joum, 
Inst. EUctr. Engrs. (British), Vol. 40 (1907-8), p. 568. E. Arnold in his 
Wechselstromiechnikf Vol. 5, Part 1, pp. 42, 43, calculates the value of ka 
for an induction machine in a somewhat different way. With open slots 
in the stator, Arnold's method gives lower values of ka than they are in reality. 
See Hoock and Hellmund, Beitrag zur Berechnung des Magnetizierungs- 



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98 THE MAGNETIC CIRCUIT [Art. 37 

Referring to Art. 36, eq. (50) may be interpreted as follows: 
Let ka be the air-gap factor for the slotted stator and a smooth- 
body rotor; let V be the same factor for the slotted rotor and a 
smooth-body stator. Then the air-gap factor of the actual 
machine 

*a=VxA;a" (51) 

In an induction motor the magnetic flux is distributed in the 
air-gap approximately according to the sine law, due to the dis- 
tributed polyphase windings. Therefore the value of M deter- 
mined from eq. (44) gives only the average value of the m.m.f. 
required for the air-gap. With a sine-wave distribution of the 
flux the maximum m.m.f. is V2 times larger than the average 
value. 

Prob. 20. What is the permeance of the air-gap of a 16-pole direct- 
current machine, the armature of which has a diameter of 250 cm. and 
is provided with 324 slots, 12 by 15 mm. ? The gross length of the armature 
is 23 cm., and it is provided with three ventilating ducts, 10 mm. wide 
each. The axial length of the poles is 21.5 cm. The pole shoes cover 
65 per cent of the periphery, and are not chamfered. The length of the 
air-gap is 10 mm. Ans. About 900 perm. 

Prob. 21. The machine mentioned in problem 14 has 120 slots, 3 by 
6.5 cm. The pole-shoes are shaped according to the arc of a circle of a 
radius equal to 90 cm. and subtending 36 degrees ; the pole-tips are formed 
by quadrants of a radius equal to 2.5 cm. Check the value of the field 
current (52 amp.) given in problem 15, by the method of equivalent 
permeances. 

Prob. 22. What is the maximum m.m.f. across the air-gap of an 
induction motor, if the gross average flux density in the air-gap (total 
flux divided by the gross area of the air-gap, not including the vents) 
is 3 kl./sq.cm., and the clearance is 1.2 mm.? The bore is 64 cm.; the 
stator is provided with 48 open slots, 22 by 43 mm. The rotor has 91 
half-closed slots, the slot opening being 3 mm. The machine has a vent 
7 mm. wide for every 9 cm. of the laminations. 

Ans. 820 amp. turns. 

Prob. 23. Show that J«/a = 1.2+2.93 log (s/2a), if the fringing 
lines of force are assumed to be concentric quadrants (Fig. 27, to the left) 
with the points c as the center; the average length of path in the part 
hccf is estimated to be equal to 1.2a, and the average width 0.72a. Hint: 
The permeance of an infinitesimal tube of force of a radius x and of a 
width dx is fi-dx/iinx). Integrate this expression between the limits 

stromas in Induktionsmotoren, Elektrotechnik und Maschinenbau, Vol. 28 
(1910), p. 743. 



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Chap. V] 



EXCITING AMPERE-TURNS 



99 



of a and Js. See Adams, he. cU., p. 332. The formula can be used only 
when s is larger than 2a. 

Prob. 24. Show that J t /a =2.93 log (1 +i?rs/a), if the fringing lines 
of force are assumed to consist of concentric quadrants (Fig. 27, to 





Fig. 27.— -Two simplified paths for the fringing flux. 

the right), with the point c' as a center, continued as straight lines. 
See Arnold, Die Gleichstrommachinej Vol. 1, p. 269. 

Prob. 26. Show that formula (50) applies to synchronous and direct- 
current machines with salient poles as well, if ^2' is the width of the 
pole shoe, corrected for the fringe, and A, is the pole pitch. 



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CHAPTER VI 

EXCITING AMPERE-TURNS IN ELECTRICAL 

MACHINERY— (C(?«^mM^d) 

38. TheAmpere-turnsRequired for Saturated Teeth. The teeth 
and the slots of an armature, under the poles, are magnetically in 
parallel (Fig. 24) ; hence, part of the flux passes from the pole into 
the armature core through the slots between the teeth. But, with 
a moderate saturation in the teeth, say below 18 kilolines per 
square centimeter, the amoimt of the flux which passes through the 
slots is altogether negligible. If the taper of the teeth is slight, 
the required ampere-turns are foimd for the average flux density 
in the tooth, taking the value of H from the curves in Fig. 3. 

Should the taper of the teeth be considerable, as is the case in 
revolving armatures of small diameter, the flux density should be 
determined in say three places along the tooth, viz., at the root, 
in the middle part, and at the crown. Let the corresponding 
values of magnetic intensity from the magnetization curve of the 
material be Hq, Hm, and Hi. Assuming H to vary along the tooth 
according to a parabolic law, we have, according to Simpson's 
rule in the first approximation, that the average intensity over the 
tooth is 

Have^iHo+iHfnHHl (52) 

If a greater accuracy is desired, the values of H can be determined 
for more than three cross-sections of the tooth and Simpson's 
rule applied.^ For instance, let the length be divided into n equal 

* A designer who has to calculate ampere-turns for teeth frequently 
will save time by plotting curves for the average H against the flux density 
Bq at the root of the teeth. Each curve would be for one taper, and these 
curves would cover the usual range of taper in the teeth. See A. Miller Gray 
''Magnetomotive Force in Non-uniform Magnetic Paths," Electrical World, 
Vol. 57 (1911), p. 111. 

100 



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Chap. VI] EXCITING AMPERE-TURNS 101 

parts, where n is an even number. Then, we have that 

-" 6n 

+ 2(^2 + ^4+... + ffn-2)]. . (53) 

When the flux density in the teeth is considerable, say between 
18 and 24 kilomaxwells per square centimeter, an appreciable part 
of the total flux passes through the slots between the teeth, also 
through the air-ducts, and in the insulation between the lamina- 
tions. Dividing, therefore, the flux per tooth pitch by the net 
cross-section of the tooth, one gets only the so-called apparent flux 
density in the tooth, which density is higher than the true density. 
With highly saturated teeth, a small difference in the estimated 
flux density makes an appreciable difference in the required number 
of ampere-turns; it is therefore of importance to know how to 
determine the true density in a tooth, knowing the apparent 
density. 

Consider first the case of a machine with a large diameter, in 
which the taper of the teeth can be neglected. Assume the con- 
centric cylindrical surfaces at the tips and at the roots of the teeth 
to be equipotential surfaces, and the lines of force*to be all parallel 
to each other, in the slots as well as in the iron. In reality, some 
lines of force enter the teeth on the sides of the slots (Fig. 24), so 
that the foregoing assumptions are not quite correct; but they are 
the simplest ones that can be made. Any other assumptions 
would lead to calculations too complicated for practical use. 

Let Bj.^1 be the true flux density in the iron of the tooth, and 
let 6app be the apparent flux density in the tooth imder the assump- 
tion that no flux passes through the slots, air-ducts, or insulation 
between the laminations. Then, denoting the actual flux density 
in the air by Bay we have the following expression for the total 
flux per tooth pitch: 

where Ai and Aa are the cross-sections in square centimeters of the 
paths per tooth pitch, in the iron and air respectively. Since the 
iron and the air paths are of equal length, and are in parallel, the 
m.m.f . •gradient is the same in both. Let H be this gradient, in 



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102 THE MAGNETIC CIRCUIT [Art. 38 

kiloampere-tums per centimeter. Then, if all the flux densities 
are in kilomaxwells per square centimeter, 

Substituting this value of Ba into the preceding equation we 
obtain, after division by Ai, 

Bapp-^B^^+l,25{Aa/Ai)H (54) 

The ratio AJAi can be expressed through the dimensions of 
the machine as follows: Ai^Un where t is the width of the tooth, 
and In is the net axial length of the laminations, without the air- 
ducts and insulation. Aa^Ug—Un, where X is the tooth pitch 
(Fig. 20), and Ig is the gross length of the armature core. Hence, 

Aa/Ai^iX/Odg/ln)-! (55) 

In eq. (54) the flux density Bapp and the ratio AJAi are known 
in any particular case, and the problem is to find B^^ and H. 
The other equation which connects B^^ and H is the magnetiza- 
tion curve of the material, and the problem can be solved in a simi- 
lar manner to problem 11 in chapter II (see also problem 4 below). 

Professional designers use curves like those shown in Fig. 28, 
which give directly the relation between Bapp and Breai within the 
range of values of AJAi which occur in practice. The curves are 
plotted point by point by assiuning certain values of B^^i and cal- 
culating the corresponding Bapp from eq. (54). For instance, for 
-8,.^= 24, the saturation curve shown in Fig. 28 gives H= 1.33, so 
that for Aa/Ai=-2, we have: 5app = 24 + 1.25X2X1.33 = 27.33. 
This determines one point on the curve marked " Ratio of air to 
iron = 2." In using these curves one begins with the known value 
of Bapp on the lower axis of abscissae, and follows the ordinate to 
the intersection with the curve for the desired ratio Aa/Ai; this 
gives the value of B^^. By following the horizontal line from the 
point so located to the intersection with the B-H curve, the corre- 
sponding value oi H is read off on the upper axis of abscissae. 

The curves in Fig. 28 are completely determined by the shape 
of the B-H curve, so that, if the material to be used for the arma- 
ture core differs considerably from that assumed in Fig. 28, new 
curves of B^.^ versus Bapp ought to be plotted, or else the method 



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Chap. VI] 



EXCITING AMPERE-TURNS 



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104 THE MAGNETIC CIRCUIT [Art. 38 

may be used which is suggested in problem 4 below. A comparison 
of the B'H curve with those in Fig. 3 shows that a much better 
quality of steel is presupposed in Fig. 28. Such is usually the case 
when it is desired to employ highly saturated teeth, for otherwise 
it might be practically impossible to get the required flux. The 
curves in Fig. 3 refer to an average quality of electrical steel. 

Formula (54) and the curves in Fig. 28 presuppose that the 
teeth have no taper, or that the taper is negligible. If the taper of 
the teeth is quite considerable the tooth and the slot. are divided 
by equipotential cylindrical surfaces into two or more parts, and 
H is determined separately for each part. Then the effective value 
of H is calculated according to Simpson's rule, using either formula 
(52) or (53). 

Prob. 1. A four-pole direct-current armature has the following 
dimensions: diameter 45 cm.; gross length of core 20 cm.; two air-ducts 
7 mm. each; 67 open slots 1 by 3 cm. The poles are of such a shape 
that the flux per pole is carried miiformly by 11.5 teeth. How many 
ampere-tums per pole are required for the teeth when the flux per pole 
is 3 megalines? Use the saturation curve for carbon-steel laminations 
in Fig. 3. Ans. 148 amp.-tums. 

Prob. 2. How many ampere-tums are required in the preceding 
problem when the flux per pole is 4.4 megalines? 

Ans. Between 2400 and 2500. 

Prob. 3. The machine, in problem 22 of the preceding chapter, had 
a gross average flux density in the air-gap of 3 kl./sq, cm. The bore 
was 64 cm. The stator was provided with 48 slots 22 by 43 mm. The 
machine has a vent 7 mm. wide for every 9 cm. of the laminations. 
What is the maximum m.m.f . required for the stator and rotor teeth, if 
the size of each of the 91 rotor slots is 14 by 30 mm. below the overhang? 
Ans. between 2400 and 2500 amp.-turns. 

Prob. 4. Instead of drawing the curves shown in Fig. 28, the relation 
between BrcaZ and Bapp can be found by the following construction: 
Disregard the lower scale marked " Apparent Flux Density "; extend 
the left-hand scale to the division 34 and mark the scale " Real and Appar- 
ent flux density." Cut out a strip of paper, and copy the left-hand scale 
on the left-hand edge of the strip. On the right-hand edge of the strip 
mark the scale for AalAi as follows: division 26 of the flux density to 
correspond with zero, division 27 with 0.4, division 28 with 0.8, etc. 
Apply the left-hand edge of the strip to division 2.0 on the upper horizon- 
tal scale, and to division 26 on the lower horizontal scale. Move the 
strip up and down until the upper horizontal scale coincides with the 
desired value of Aa/At marked on the strip. Lay a straightedge on the 
divisions of the two vertical scales corresponding to the given apparent 
flux density. The intersection of the straightedge with the B-H curve 



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Chap. VI] EXCITING AMPERE-TURNS 105 

will give the required values of Breai and H, Check this construction 
for a few points with the values obtained from the curves, and give a 
general proof. Hint: This construction amounts to considering the 
B-H curve and Eq. (54) as two simultaneous equations with two unknown 
quantities Breai and H, See problem 11 in chapter II, Art. 13. 

39. The Ampere-ttims for the Armature Core and for the 
Field Frame. In many machines the m.m.f. required for the air- 
gap and the teeth are large as compared to those required for the 
armature core and the field frame; in such cases the latter are 
either altogether neglected, or are estimated roughly, by increas- 
ing the ampere-turns calculated for the rest of the magnetic cir- 
cuit by say five or ten per cent. Where this is not permissible, the 
usual procedure is to estimate the maximum flux density in the 
core or frame under consideration and to measure from the drawing 
of the machine the length of the average path of the lines of force 
in it. The assumption is made that the same flux density is main- 
tained on the whole length of the path, and the required ampere- 
turns are calculated from the magnetization curve of the material 
(Figs. 2 and 3). While the ampere-turns determined in this way 
are usually larger than those actually required, the method is per- 
missible if the total amount of the m.m.f . for the parts under con- 
sideration is small as compared to the total m.m.f. of the magnetic 
circuit. If a greater accuracy is desired, the path is subdivided 
into two or more parts in series, and the average density deter- 
mined for each part; and then the ampere-turns required for each 
part are added. 

The tendency now is to increase the flux density in the arma- 
ture cores of alternators and induction motors so as to reduce the 
size of the machine. This is made possible through a better quality 
of laminations, which show a smaller core loss, and also through the 
use of a more intensive ventilation. With these high densities 
and with the comparative large values of the pole pitch necessary in 
high-speed machinery, the ampere-turns for the core constitute an 
appreciable amount of the total m.m.f. of the machine, and it is 
therefore desirable to calculate them more accurately. 

The flux density in the core is a minimum opposite the center; 
of a pole, and is a maximum in the radial plane midway between ' 
two poles (Fig. 15). At each point the flux density has a tangen- 
tial and a radial component. The latter is comparatively small 
and can be neglected; the tangential component can be assumed 



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106 THE MAGNETIC CIRCUIT [Aet. 39 

to vary according to the sine law, being zero opposite the center of 
the pole and reaching its maximum between the poles. With 
these assumptions, knowing the maximum flux density in the core, 
the flux density at all other points is calculated, and the corre- 
sponding values of H are determined from the B-H curve of the 
material. The average value of H for one-half pole pitch is then 
found by Simpson's rule, eqs. (52) and (53). With the sine-wave 
assumption, the average H depends only upon the maximum 
flux density, so that for a given material a curve can be compiled 
from the B-H curve, giving directly Have for different values of 
B 1 

Should a still greater accuracy be required, the following 
method can be used : Draw the assumed or the calculated ciu-ve 
of the distribution of flux density in the air-gap, and indicate to 
your best judgment the tubes of force in the armature core, say 
for each tooth pitch. The flux in the radial plane midway between 
the two poles can be assumed to be distributed uniformly over the 
cross-section, and this fact facilitates greatly the determination of 
the shape of the tubes of flux. The m.m.f . required for each tube 
is calculated by dividing it into smaller tubes in series and in paral- 
lel; thus, either the average m.m.f. for the whole flux can be found, 
or the maximum m.m.f. for one particular tube.^ 

The frame to which the poles are fastened in direct-current and 
in synchronous machines is usually made of cast iron; in some 
cases the frame is made of cast steel; in high-speed synchronous 
machines the revolving field is made of forged steel. The magneto- 
motive force required for such a frame is f oimd in the usual way 
from the magnetization curve of the material, knowing the area 
and the average length of the path between two poles; the length 
is estimated from the drawing of the machine. In figuring out the 
flux density in a field frame one must not forget that (1) only one- 
half of the flux per pole passes through a given cross-section of the 
frame (Fig. 20) ; (2) the total flux in the frame and in the poles is 
larger than that in the armature by the amoimt of the leakage flux 
between the poles. This leakage is usually estimated in per cent 

* This method is due to E. Arnold. See his WechaeUtromtechnik, Vol. 5, 
(1909), part 1, p. 48. 

2 For details of this method see Hoock and Hellmund, Beitrag zur Berech- 
nung des Magnetizienmgsstromes in Induktionsmotoren, Elektrotechnik und 
Maschinenbau, Vol. 28 (1910), p. 743. 



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Chap. VI] EXCITING>MPERE.TURNS 107 

of the useful flux, from one's experience with previously buUt 
machines, or it can be calculated by the methods explained in the 
next article. Thus, a leakage factor of 1.20 means that the flux in 
the field poles is 20 per cent higher than that in the armature, the 
leakage flux constituting 20 per cent of the useful flux. The usual 
values of the leakage factor vary between 1.10 and 1.25, depending 
upon the proximity of the adjacent poles, the degree of saturation 
of the circuit, and the proportions of the machine. 

The ampere-turns required for the pole-pieces are calculated in 
a similar way, assuming the whole leakage to take place between 
the pole-tips, so that the flux density in the pole-waist corresponds 
to the total flux, including the leakage flux. In exceptional cases 
of highly saturated pole-cores this method may be inadmissible, on 
account of too large a margin which it would give as compared to 
the ampere-turns actually required. In such cases part of the 
leakage may be assumed to be concentrated between some two 
corresponding points on the waists of two adjacent poles, or it may 
be assumed to be actually distributed between the two pole-waists. 
See probs. 9 and 10 in chapter II. 

In some machines the joint between the pole and the frame 
offers a perceptible reluctance, like the joints in the transformer 
cores discussed in Art. 33. Some designers allow a certain 
fraction of a millimeter of air-gap to account for this reluctance, 
and add the number of ampere-turns required to maintain the 
flux in this air-gap to those for the pole-piece. The length of 
this equivalent air-gap is found by checking back no-load 
saturation curves obtained from experiment. As a usual rule, 
it is advisable to increase the total calculated ampere-turns of the 
magnetic circuit by about 5 to .10 per cent. This increase covers 
such minor points as the reluctance of the joints, omitted in 
calculations, as well as certain inaccurate assmnptions; it also 
covers a possible discrepancy between the assumed and the 
actual permeability of the iron. With a liberally proportioned 
field winding and a proper regulating rheostat a designer can 
rest assured that the required voltage will be obtained, though 
possibly at a somewhat different value of the field current than 
the estimated one. 

Prob. 6. The stater core of a six-pole induction motor has the 
following dimensions: bore 112 cm.; outside diameter 145 cm.; gross 
length 55 cm. ; the slots are 2 cm. X4.5 cm. ; the machine is provided with 



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108 THE MAGNETIC CIRCUIT [Art. 40 

8 ventilating ducts 9 mm. wide each. What is the maximum m.mi. 
required for the stator core per pole if the flux per pole is 0.15 weber? 

Ans. 1 90 using Arnold 's method. 

Prob. 6. Draw a curve between the average H and maximum B 
in the core, assuming a sinusoidal distribution of the flux density in the 
tangential direction, for the carbon steel laminations in Fig. 3. 

Ans. Have = 26.5 for Bmax = 18. 

Prob. 7. The cross-section of the cast-iron field yoke of a direct- 
current machine is 370 sq.cm.; the mean length of path in it between 
two consecutive poles is 85 cm. The length of the hnes of force in each 
pole-waist is 21 cm.; its cross-section 420 sq.cm. The poles are made 
of steel laminations 4 mm. thick, so that the space lost between the 
laminations is negligible. The reluctance of the joint between a bolted 
pole and the yoke is estimated to be equivalent to 0.1 mm. of air. What 
is the required number of ampere-turns for the pole-piece and the yoke, 
per pole, when the useful flux of the machine is 5 megalines per pole? 
The leakage factor is estimated to be equal to 1.20. 

Ans. About 930. 

40. Magnetic Leakage between Field Poles. It is of impor- 
tance in modem highly saturated machines to know accurately the 
leakage flux between the poles, in order to estimate correctly the 
ampere-turns required for the field poles and the frame of the 
machine. Moreover, the design of the poles can be improved, 
knowing exactly where the principal leakage occurs and how it 
depends upon the proportions of the machine. The value of the 
leakage factor also affects the voltage regulation of the machine, 
because at full load the m.m.f. between the pole-tips has to be 
larger than at no-load, on account of the armature reaction. 

For new machines of usual proportions the value of the leakage 
factor can be estimated from tests made upon similar machines. 
But in new machines of unusual proportions the designer has to 
rely upon his judgment, assisted if necessary by crude compara- 
tive computations of the permeance between adjacent poles. In 
this and in the next article some examples of such computations 
are given, not so much in order to give a definite method to be fol- 
lowed in all cases, as to show the student a possible procedure and 
to train his judgment in estimating the permeance of an irregular 
path. 

Four principal paths of leakage can be distinguished between 
two adjacent poles (Fig. 29) : (a) between the sides of the pole 
shoes which face each other; (b) between the sides of the pole 
cores (waists) parallel to the shaft of the machine; (c) between the 



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Chap. VI] 



EXCITING AMPERE-TURNS 



109 



flanks or sides of the pole shoes perpendicular to the shaft; (d) 
between the flanks of the pole cores. In the calculations which 
follow, the permeances are computed between a pole and the planes 
of symmetry, MN, between the two poles, the permeance of the 
other half of each path being the same. All these leakage paths 
are in parallel with respect to the pole, so that the total leakage 




Fig. 29. — ^The leakage flux between field poles. 

permeance is equal to their sum. Kaowing this total permeance 
and the m.m.f. between the pole and the plane MN the leakage 
flux is found, and knowing this flux and the useful flux per pole 
the leakage factor is easily calculated. 

We shall now estimate the permeances of each of the four above 
mentioned paths of leakage. 

(a) Between the adjacent pole-tips. Estimate the average 
cross-section A of the path, in square centimeters, and the aver- 



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110 THE MAGNETIC CIRCUIT [Art. 40 

age length I between the pole-tip and the plane MN, being guided 
by Fig. 29. (Do not encroach upon the fringe to the armature.) 
Then the permeance of the path is, in perms, 

(P=^fiA/l (56) 

If a greater accuracy is desired, subdivide the total path into 
smaller paths in series and in parallel, and calculate the permeance 
or the reluctance of each separately. Then the total permeance is 
foimd according to the well-known law of combination of reluc- 
tances and permeances in series and in parallel (Art. 9). When 
mapping out the lines of force in the air, begin them nearly at right 
angles to the surface of the pole (see Art. 41 a below) and draw 
them so as to make the total permeance of the path a maximum, 
that is, reducing as far as possible the length and increasing the 
cross-section of each elementary tube of flux. The medium may 
be said to be in a state of tension along the lines of force, and of 
compression at right angles to their direction, by virtue of the 
energy stored in the field. Hence, there is a tendency for the tubes 
of force to contract along their length and expand across their 
width. 

(b) Between the opposite pole-cores. In this part of the leakage 
field each elementary concentric path is subjected to a different 
m.m.f., that between the roots of the poles being practically zero, 
while the m.m.f. between the points p and />' is equal to that 
between the pole-tips. In most cases it is permissible to consider 
the whole leakage flux as if passing through the whole length of the 
pole core, and then crossing to the adjacent poles at the pole-tips. 
Therefore, it is convenient to add the permeance between the pole 
cores to that between the pole-tips. But the average m.m.f. 
between the waists is only about one-half of that between the tips, 
so that the equivalent permeance between the pole cores, reduced 
to the total m.m.f., is equal to one-half of the actual permeance. 
If the actual permeance, calculated according to formula (56) is(P, 
the effective permeance is i(P. The average length and cross- 
section of the path are easily estimated from the drawing of the 
machine. 

(c) Between thejlanks of the pole shoes. The path extends indefi- 
nitely outside the machine, and the lines of force are twisted 
curves, so that it is diflScult to estimate the permeance graphically. 
As a rough estimate, this permeance can be reduced to that of the 



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Chap. VI] 



EXCITING AMPERE-TURNS 



111 



path in the air between two rectangular poles of an electromagnet 
(Fig. 30). Assume the paths of the flux to consist of concentric 
quadrants with the centers at c and c', joined by parallel straight 
lines, and let the width of the poles in the direction perpendicular 
to the plane of the paper be A. Then the permeance of an infin- 
itesimal layer of thickness dx, between one of the poles and the 
plane MN of symmetry, is 

d(P=fdx.h/(i7:x+l). 

Integrating this expression between the limits o and b we find 



(P=1.84Alog (1.576/Z + l) perms 



(57) 



(compare with prob. 24 in Chapter V, Art. 37). 

In applying this formula and Fig. 30 to the case of the flank 
leakage between the pole shoes, h is the average radial height of 
the pole shoe, 6 is equal to one- 
half the width of the pole shoe, 
and 2Z is the distance between 
the two opposing pole-tips. 
While the method evidently 
gives only a crude approxi- 
mation to the actual perme- 
ance, formula (57) at least 
fixes a lower limit to the per- 
meance in question. 

(d) Between the flanks of 
the pole cores. The conditions 
are similar to those imder (c), 
so that the permeance is esti- 
mated again on the basis of 
formula (57) . The sides of the 
two rectangles in Fig. 29 are 
not parallel to each other as in 
Fig. 30, but this difference is 
taken into account by mentally 
turning^ them into a parallel 
position, and estimating the equivalent distance 21 between the 
edges of the opposing poles. The dimension h is in this case the 
radial height of the pole-waist, and b is one-half of the width of 
the pole-waist. The flank leakage is smaller than that between 




Fig. 30. — ^The magnetic path between 
the poles of an electromagnet. 



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112 THE MAGNETIC CIRCUIT [Art. 40 

the opposite side, so that one may be satisfied with a lesser degree 
of accuracy. The equivalent permeance, reduced to that between 
the pole-tips, is again equal to one-half the actual permeance, for 
the same reason as imder (6) above. 

The total leakage permeance between a pole and the two planes 
of symmetry is equal to the sum of the permeances calculated as 
above. In summing them up it will be seen from Fig. 29 that the 
. permeaoces (a) and (b) must be taken twice, and also that (c) and 
(d) must be taken four times. The leakage flux per pole is 
obtained by multiplying the total leakage permeance by the m.m.f . 
between the pole-tip and the plane of symmetry. This m.m.f . is 
equal to that required to establish the useful flux, along the path 
grs, through the air-gap and the armature of the machine, and 
consequently it is known before the pole-piece and the field wind- 
ing are computed in detail. Knowing the leakage flux and the 
useful flux, the leakage factor is figured out according to the defi- 
nition given above. 

When calculating permeances as indicated above, one is advised 
to make liberal estimates of the same, for two reasons: In the first 
place, the true permeance of a path is always the largest possible, so 
that, whatever assumptions one makes, the calculated permeance 
comes out smaller than the actual. In the second place, in design- 
ing a new machine it is better to be on the safe side and rather 
underestimate than overestimate the excellence of the perform- 
ance. Some writers give more elaborate rules and formulae for the 
calculation of the leakage permeance which are usefiQ in the 
derign of machines of special importance.^ 

The leakage factor remains practically constant as long as the 
flux density in the armature core and teeth is moderate, so that 
the reluctance of the useful path grs is nearly constaot. This is 
because the reluctance of the leakage paths is constant, and, if the 
reluctance of the useful path is also constant, the useful flux and 
the leakage flux increase in the same proportion when the m.m.f. 
between the pole-tips is increased. When the armature iron is 
approaching saturation, the leakage factor increases with the field 

^ For a more detailed treatment of the leakage between poles see the 
following works: E. Arnold, Die Oleichstrommaschiney Vol. 1 (1906), pp. 
284-294; Hawkins and WaUis, The Dynamo, Vol. 1 (1909), pp. 469-484; 
Pichelmayer, Dynamohau (1908), pp. 127-131; Cramp, CorUimums-Current 
Maehine Design (1910), pp. 42-47 and 226-230. 



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Chap. VI] EXCITING AMPERE-TURNS 113 

current, because the leakage flux increases the more rapidly than 
the useful flux. This increase is partly offset by the fact that the 
pole-tips also become gradually saturated by the leakage flux, so 
that the leakage factor does not increase as rapidly as it would 
otherwise. The practical point to be observed is, that for the 
higher flux densities, if accuracy is desired, the leakage should 
be estimated separately for a few points on the no-load saturation 
curve. 

For a given terminal voltage, the leakage factor of a machine is 
somewhat higher at full-load than at no-load, because the required 
m.m.f . between the pole-faces is higher, due to the armature reac- 
tion and to the voltage drop in the armature. In comparatively 
rare cases, when the armature reaction assists the field m.m.f ., for 
instance, in the case of an alternator supplying a leading current, 
the leakage factor decreases with the increasing load. The fol- 
lowing example illustrates the influence of the load upon the value 
of the leakage factor. 

Let the useful flux per pole in an alternator, at the rated 
voltage and at no-load, be 5 megalines, and let 6000 amp. -turns 
per pole be required for the air-gap and the armature core. Let 
the permeance of the leakage paths between a pole and the neutral 
planes be 120 perms, so that the leakage flux is 0.72 megaline, and 
the leakage factor is (5.00 +0.72)/5.00= 1.14. Let a useful flux 
of 5.5 megalines be required at the same voltage and at full load, 
an increase of 10 per cent being necessary to compensate for the 
internal drop of voltage due to the armature impedance. If the 
teeth and the armature core were not saturated at all, an m.m.f. 
of 6600 amp.-tums would be required. Li reality, the m.m.f. is 
higher, say 7500 amp.-tums. Let the armature reaction be equal 
to 1500 demagnetizing ampere-turns per pole. To compensate 
for its action, 1500 additional ampere-turns are required on each 
field coil. Thus, the difference of magnetic potential between 
a pole-tip and the adjacent plane of symmetry MN (Fig. 29) 
is now 9000 amp.-tums, and the leakage flux is increased to 
1.08 megalines. Therefore, the leakage factor at full load is 
(5.50 + 1.08)/5.50 = 1.20. Similar relations hold for the direct- 
current machines. 

In calculating the performance of a synchronous or a direct- 
current machine one has to use the relation between the field cur- 
rent and the voltage induced in the armature. Ordinarily, the 



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114 THE MAGNETIC CIRCUIT [Art. 40 

no-load saturation curve is used for this purpose, assuming that 
the leakage factor is the same at full load as at no load. However 
careful designers sometimes plot a separate curve, using a higher 
leakage factor, for use at full load. 

Prob. 8. Assume in the illustrative example given in the text the 
armature current to be leading, so that the voltage drop in the armature 
is negative and the armature reaction strengthens the field. Show that 
with the same value of the armature current the leakage factor is about 
1.09. 

Prob. 9. Draw rough sketches of the magnetic circuits of two 
machines, one possessing such proportions, number and shape of poles 
as to give a particularly low leakage factor, the other markedly deficient 
in this respect. 

Prob. 10. Calculate the leakage factor and the leakage permeance 
per pole of a six-pole turbo-alternator of the following dimensions; the 
bore is 1.2 m.; the axial length of the poles 0.6 m.; minimum air-gap 
1 cm.; maximum air-gap 2 cm.; total height of the pole 23 cm.; the 
height of the pole-waist 18 cm.; the breadth across the pole-waist 25 cm.; 
that across the pole-tips 36 cm. The reluctance of the useful path in 
the air-gap and in the armature is estimated to be about 0.57 millirel 
per pole. Ans. 1.115; about 200 perms. 

Prob. 11. The leakage factor of the machine specified in the pre- 
ceding problem was found from an experiment to be 1.13, at no-load, 
when the total flux per pole was 20.35 megalines. What is the true 
leakage permeance if 20 kiloampere-turns were required at that flux for 
the air-gaps and the armature, per pair of poles ? Ans. 234 perms. 

Prob. 12. The machine specified in the two foregoing problems 
requires at full load 20 per cent more ampere-turns for the air-gap and 
armature, on account of the induced voltage being 12 per cent higher 
than at no-load. The armature reaction amounts to 4000 demagnetizing 
ampere-turns per pole. What is the leakage factor at full load, according 
to the calculated leakage permeance and according to that obtained 
from the test? Ans. 1.16; 1.19. 

Prob. 13. A closed electric circuit consisting of a battery and of 
a bare conductor is immersed in a slightly conducting liquid, so that part 
of the current flows through the liquid. Indicate the common points 
and the difference between this arrangement and a magnetic circuit 
with leakage. Using the electrical analogy, show that armature reaction 
increases the leakage factor; also explain the fact that, in order to com- 
pensate for the action of M demagnetizing ampere-turns on the armature, 
more than M additional ampere-turns are required on the pole-pieces. 

Prob. 14. In some books the permeance between two pole-faces 
(Fig. 30) is calculated by assuming the lines of force to be concentric 
semicircles as shown by the dotted lines. Show that such a permeance 
is smaller than that according to formula (57) and therefore should not 
be used. Hint: Compare the lengths of two corresponding lines of 
force. 



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Chap. VI] 



EXCITING AMPERE-TURNS 



115 



Prob. 15. Let AB and CD (Fig. 31) represent the cross-sections 
of two opposite pole-faces of an electromagnet, inclined at an angle 20 
to one another. Show that of the three assumptions with regard to the 
shape of the Hnes of force in the air between the poles (a) is more correct 
than (6) and (6) is more correct than (c) ; in other words, the assumption 
(a) gives a higher permeance than (6) or (c). Hint: tan ^ > ^ > sin ^. 

Prob. 16. Show that the permeance according to Fig. 31a, between 
one of the faces and the plane MN of symmetry, is equal to 

{fih/d)Ln{Ow/l + l), 

where h is the width of the pole-faces perpendicular to the plane of the 
paper, and that formulae (56) and (57) are special cases of it. 

Prob. 17. The formula given in the preceding problem is deduced 
under the assumption that the same m.m.f. is acting on all the lines of 





(c) 



Fig. 31. — ^The magnetic paths between the poles of an electromagnet (three 

assumptions). 

force. Let now Fig. 31a represent a cross-section of two opposing pole 
cores in an electric machine, the m.m.f. between A and C being zero, 
and uniformly increasing to a value M between the points B and D, 
Show that the equivalent permeance of the path, referred to the m.m.f. 
M is equal to {fih/d)[l-(l/dw)Ln{dw/l-hl)l 

Note. If it is desired to use regularly the foregoing formula in esti- 
mating the leakage factor, the values of the expression in the brackets 
[ ] can be plotted as a curve for the values of (l/Ow) as abscissae. Similar 
formulae can be deduced and curves plotted for the permeance of the 
flank leakage between adjacent poles. The paths of the lines of force 
over the poles can be assumed to be concentric quadrants and between 
the poles to have a shape similar to that indicated in Fig. 31a. 

41. The Penneance and Reluctance of Irregular Paths. In 

using the methods described above for the calculation of the 
ampere-turns for the air-gap, the teeth, and the cores, and in esti- 



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116 THE MAGNETIC CIRCUIT [Art. 41 

mating the leakage factor, the reader has seen the diflacnlties 
involved in the computation of the permeance of an irregular path. 

In the parts of a magnetic field not occupied by the exciting 
windings, the general principle applies that the lines of force and 
the equipotential surfaces assume such shapes and directions that 
the total permeance becomes a maximum, or the reluctance a 
minimum. When this condition is fulfilled, the energy of the 
magnetic field becomes a maximum, as is explained in Art. 57. 

When the field needs to be considered in two dimensions only, 
that is, in the case where we have long cylindrical surfaces the 
properties of conjugate fimctions can be used for determining the 
equations of the lines of force and of the equipotential surfaces; 
see the refere^ces in Art. 37 above. However, the purely mathe- 
matical difficulties of the method are such as to make the analytical 
calculation of permeances feasible in the simplest cases only. 

In most practical cases, especially in three-dimensional prob- 
lems, recourse must be had to the graphical method of trial and 
approximation, in order to obtain the maximum permeance. 
The field is mapped out into small cells by means of lines of force 
and equipotential surfaces, drawing them to the best of one's 
judgment; the total permeance is calculated by properly com- 
bining the permeances of the cells in series and in parallel. Then 
the assumed directions are somewhat modified, and the permeance 
is calculated again, etc., until by successive trials the positions of 
the lines of force are found with which the permeance becomes a 
maximum. 

The work of trials is made more systematic by following a pro- 
cedure suggested by Lord Rayleigh. Imagine infinitely thin sheets 
of a material of infinite permeability to be interposed at intervals 
into the field imder consideration, in positions approximately 
coinciding with the equipotential surfaces. If these sheets exactly 
coincided with some actual equipotential surfaces, the total 
permeance of the paths would not be changed, there being no 
tendency for the flux to pass along the equipotential surfaces. In 
any other position of the infinitely conducting sheets, the total 
permeance of the field is increased, because through these sheets 
the flux densities become more uniformly distributed. Moreover, 
these sheets become new equipotential surfaces of the system, 
because no m.m.f . is required to establish a flux along a path of 
infinite permeance. Thus, by drawing in the given field a system 



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Chap. VI] EXCITING AMPERE-TURNS 117 

of surfaces approximately in the directions of the true equipo- 
tential surfaces, and assuming these arbitrary surfaces to be the 
true equipotential surfaces, the true reluctance of the path is 
reduced. In other words, by calculating the reluctances of the 
laminae between the " incorrect " equipotential surfaces and addiag 
these reluctances in series, one obtains a reluctance which is lower 
than the true reluctance of the path. This gives a lower limit for 
the required reluctance (or an upper limit for the permeance) of 
the path. 

Imagine now the various tubes of force of the original field 
wrapped up in infinitely thin sheets of a material of zero permeabil- 
ity. This does not change the reluctance of the paths, because 
there are no paths between the tubes. But if these wrappings are 
not exactly in the direction of the lines of force, the reluctance of 
the field is increased, because the densities become less uniform, 
the non-permeable wrappings forcing the lines of force from their 
natural positions. Thus, by drawing in a given field a system of 
surfaces approximately in the directions of the lines of force, cal- 
culatiQg the reluctances of the individual tubes, and adding them 
in parallel, a reluctance is obtained which is higher than the true 
reluctance of the path. This gives an upper limit for the reluc- 
tance (or a lower limit for the permeance) of the path imder 
consideration. 

Therefore, the practical procedure is as follows: Divide the 
field to the best of your judgment into cells, by equipotential 
surfaces and by tubes of force, and calculate the reluctance of the 
field in two ways: first, by adding the cells in parallel and the 
resultant laminae in series; secondly, by adding the cells in series 
and the resultant tubes in parallel. The first result is lower than 
the second. Readjust the position of the lines of force and of the 
equipotential surfaces until the two results are suflBiciently close to 
one another; an average of the two last results gives the true 
reluctance of the field. 

One difficulty in actually following out the foregoing method 
is that the changes iq the assumed directions of the field that will 
give the best result are not always obvious. Dr. Th. Lehmann has 
introduced an improvement which greatly facilitates the laying out 
of a field. ^ We shall explain this method in application to a two- 

* " Graphische Methode zur Bestimmung des Kraftlinienverlaufes in der 
Luft "; Elektrotechnische Zeitschriftj Vol. 30 (1909), p. 995. 



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118 THE MAGNETIC CIRCUIT [Art. 41 

dimensional field, though theoretically it is applicable to three- 
dimensional problems also. According to Lehmann, lines of force 
and level surfaces are drawn at such distances that they enclose 
cells of equal reluctance. Consider a slice, or a cell, in a two- 
dimensional field, V centimeters thick in the third dimension 
(where v=l/jj), and of such a form that the average length I of 
the cell in the direction of the lines of force is equal to its average 
width w in the perpendicular direction. The reluctance of such a 
cell is always equal to one rel, no matter whether the cell itself is 
large or small. This follows from the fundamental formula for 
the reluctance, which in this case becomes (R = i^l/(yXw) = l, 

The judgment of the eye helps to arrange cells of a width equal 
to the length, in the proper position with respect to each other and 
to the. adjoining iron; the next approximation is apparent from 
the diagram, by inspecting the lack of equality in the average 
width and length of the cells. Lord Rayleigh's condition is 
secured by this means, since the combination of cells of equal 
reluctance leads to but one result, whether they are combined 
first in parallel or first in series. After a few trials the space is 
properly ruled, and it simply remains to count the number of cells 
in series and in parallel. Dr. Lehmann shows a few applications 
of his method to practical cases of electrical machinery, and the 
reader is referred to the original article for further details. 

The foregoing methods apply only to the regions outside the 
exciting current, because only in such parts of the field the maxi- 
mum permeance corresponds to the maximum stored electromag- 
netic energy. Within the space occupied by the exciting windings 
the condition for the maximum of energy is different (see Art. 57), 
and is of a form which hardly permits of the convenient application 
of a graphical method. However, in most practical cases the 
directions of the lines of force within the exciting windings are 
approximately known a priori: or else, the windings themselves 
can be assumed, for the purposes of computation, to be concentrated 
within a very small space. For instance, the field winding can be 
assumed to consist of an infinitely thin layer close to the pole-waist. 
Then the condition that the permeance is a maximum is fulfilled in 
practically the whole field, and the field is mapped out on this basis. 

Prob. 18. Sketch the field between the armature and a pole-piece 
or some proportion of tooth, slot, and air-gap and determine the lower 
and upper limits of the reluctance by Lord Rayleigh's method. 



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Chap. VI] 



EXCITING AMPERE-TURNS 



119 



Prob. 19. For some ratio of slot width to air-gap draw the tooth- 
fringe field to the perpendicular surface of the pole, adjust the number 
and spacing of the lines of force by Dr. Lehmann's method, and see how 
closely you can check the corresponding point on Carter's curve (Fig. 26). 

Prob. 20. From the given drawing of a machine, determine the 
permeance of the fringe from the pole-tip to the armature by Lehmann's 
method; consult, If necessary. Dr. Lehmann's original article. 

Prob. 21. Map out the leakage field between the opposing pole-tips 
and cores of a given machine, and determine its equivalent permeance by 
Lehmann's method, assuming the field coils to be thin and close to the core. 

41a. The Law of Flux Refraction. When mapping out a field 
in air, the lines of force must be drawn so as to enter the adjoining 
iron almost normally to its surface, even if they are continued in 
the iron almost parallel to its surface. The lines of force change 
their direction at the dividing surface suddenly (Fig. 32), and in 
so doing they obey the so-called law of flux refraction; namely, 



tan Oi/tan Oa= fii/ [ia 



(58) 



EqulpotcntJiil_ 



Iron 



Since /lo is many times smaller than //t, the angle da is usually very 
small, imless Oi is very nearly 90 degrees. It may be said in gen- 
eral that the lower the permeability 
of a medium the nearer the lines 
of force are to the normal at its 
limiting surfaces. In this way, the 
path between two given points is 
shortened in the medium of lower 
permeability and is lengthened in 
the medium of higher permeability. 
Thus, the total permeance of the 
circuit is made a maximum. 

To deducef the above-stated law 
of refraction, consider a tube of 
flux between the equipotential sur- 
faces ah and cd, the width of the 
path in the direction perpendicular 
to the plane of the paper being one 
centimeter. Let Bi and Ba be the 
flux densities, and Hi and Ha the 

corresponding magnetic intensities in the two media. Two 
conditions must be satisfied, namely, first, the drop of m.m.f. 




Fig. 32.— The refraction of 
a flux. 



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120 THE MAGNETIC CIRCUIT [Akt. 41 

along ac is the same as that along M, and secondly, the total 
flux through cd is equal to that through ab. Or 

Ha • dc = Hi* bdf 

and 

Ba*cd = Bi'ab. 

Dividing one equation by the other, and rearranging the terms, 
eq. (58) is obtained. 

Prob. 22. Show that the total refraction which is in some cases 
experienced by rays of light is impossible in the case of magnetic lines 
of force. 

Prob. 23. Part of a flux emerges from the flank of a tooth into the 
slot at an angle of 1° to the normal. What is the angle which the lines 
of force make with the side of the slot in the iron, assuming the relative 
permeability of the iron to be 1000? Ans. 90°-^i=3° 17'. 



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CHAPTER VII 

THE MAGNETOMOTIVE FORCE OF DISTRIBUTED 

WINDINGS 

42. The H.H.F. of a Direct-current or Single-phase Dis- 
tributed Winding. In the two preceding chapters it is shown how 
to calculate the ampere-turns required for a given flux in an elec- 
tric machine. When the exciting winding is concentrated, that is, 
when aU the turns per pole embrace the whole flux, the number of 
ampere-turns is equal to the product of the actual amperes flowing 
Jihrough the winding times the number of turns. Such is the case 
in a transformer, in a direct-current machine, and in a sjrnchronous 
machine with salient poles. In some cases, however, the exciting 
windings are distributed along the air-gap, so that only a part of the 
flux is linked with all the turns, and the actual ampere-turns have 
to be multiplied by a factor in order to obtain the effective m.m.f . 
Such is the case in an induction motor, and in an alternator with 
non-salient poles. Moreover, one has to consider the m.m.f. of 
distributed armature windings when calculating the performance 
of a machine imder load, because the armature currents modify the 
no-load flux. In this chapter the m.m.fs. of distributed windings 
are treated mainly in application to the performance of the induc- 
tion motor; in particular, to the calculation of the no-load current 
and the reaction of the secondary currents. The armature reaction 
in synchronous and in direct-current machines is analyzed in the 
next two chapters. 

Distrilmted Winding for Alternator Field. A cross-section of a 
four-pole field structure with non-salient poles for a turbo-alterna- 
tor is shown in Fig. 33a. The flux is graded (Fig. 336) in spite of 
a constant air-gap, because the total ampere-tums act only upon 
the part a of a pole; two-thirds of the ampere-tums act upon the 
parts 6, 6 and one-third upon the parts c, c. The m.m.f . and the 
flux in the parts d, d are equal to zero. Thus, theoretically, the 
flux density in the air-gap should vary according to a " stepped '' 

121 



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122 



THE MAGNETIC CIRCUIT 



[Abt. 42 




Fig. 33a. A four-pole revolving fieldstructure with non-salient poles. 

curve (Fig. 336) ; in reality, the comers are smoothed out by the 
fringes. The total number of ampere-tiuns per pole must be such 

as to create the assumed 
maximum flux density 
Bfnax iDL the air-gap 
imder the middle part 
of the pole. The slots 
are placed with due re- 
gard to the mechanical 
strength of the struc- 

FiG. 336. The flux-density distribution for *^^' ^^.®^ ^® ^.^ Sft a 
the field shown in Fig. 33a. flux-density distribu- 

tion approaching a sine 
wave. The middle part of the curve is left flat, because very little 




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Chap. VII] M.M.F. OF DISTRIBUTED WINDINGS 



123 



flux would be gained by placing a narrow coil near the center of 
the pole, at a considerable expense in copper, and in power loss for 
excitation. The total flux, which is proportional to the area of the 
curve in Fig. 336, must be of the magnitude required by eq. (31) 
for the induced e.mi . If greater accuracy is desired, the curve in 
Fig. 336 is resolved into its fundamental sine wave and higher 
harmonics; the area of the fundamental curve must then give 
the flux which enters into eq. (31). 

Single-phdse Distributed Winding. Let us consider now the 
stator winding of an induction motor, and in particular the m.m.f . 
created by the current in one phase. We begin with the simplest 
case of a winding placed in one slot per pole per phase (Fig. 34). 
The reluctances of the 

stator core and of the ^' 

rotor core are small as 
compared with that of 
the air-gap and the 
teeth, and are taken 
into accoimt by in- 
creasing the reluctance 
of the active layer of 
the machine (air-gap 
and teeth). If Pand 
Q are the centers of 
the slots in which the 
opposite sides of a coil 
are placed, the m.m.f . 
distribution along the air-gap is that shown by the broken line 
RPP'Q'QS. In other words, the m.m.f. across the active layer, 
at any instant, is constant over a pole pitch, and is alternately 
positive and negative under consecutive poles. 

Let n be the number of turns per pole, and i the instanta- 
neous current; then the height PP^ of the rectangle is equal to ni. 
It is imderstood of course that such an m.m.f. acting alone does 
not produce the sinusoidal distribution of the flux density assumed 
in the previous chapters: In a single-phase motor the sinusoidal 
distribution is due to the simultaneous action of the stator and 
rotor currents, and also to the fact that the windings are distrib- 
uted in several slots per pole. In a polyphase machine the simul- 
taneous action of the two or three phases also helps to secure a 




Fig. 34. — ^The m.m.f. of a sbgle-phase unislot 
winding resolved into its harmonics. 



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124 THE MAGNETIC CIRCUIT [Art. 42 

sinusoidal distribution. As long as the coil PQ acts alone, the 
m.m.f. has a " rectangular " distribution in space, and, if the cur- 
rent in the coil varies with the time according to the sine law, the 
height of the rectangle, or the m.m.f. across the active layer, also 
varies according to the sine law. In what follows it is important 
to distinguish between variations of the m.m.fs. in space, i.e., along 
the air-gap, and those occurring in time, as the current in a winding 
varies. 

For the purposes of analysis the rectangular distribution of the 
m.m.f . can be replaced by an infinite number of sinusoidal distribu- 
tions (Fig. 34), according to Fourier's series.^ The advantages of 
such a development over the orginal rectangle PP^Q^Q are as fol- 
lows: 

(a) The sine wave is a familiar standard by which all other 
shapes of periodic cunres are judged. 

(b) When adding the m.m.fs. due to the coils in different slots, 
or belonging to different phases, it is much more convenient to add 
sine waves than to add rectangles displaced in space and varying 
with the time. 

(c) In the actual operation of an induction motor or generator 
the higher harmonics in the m.m.f. wave are to a considerable 
extent wiped out by the corresponding currents in the rotor, so that 
the rectangular distribution is actually changed to a nearly sinu- 
soidal one (see Art. 45 below). 

Let h be the height of the rectangle; we assume that for all the 
points along the air-gap the sum of the ordinates of all the sine 
waves is equal to A; or 

A = ili sinx+^a sinSx+^ssin Sx+etc. . . (59) 

Here x is the angle in electrical degrees, counted along the air-gap, 
and Aij Az, As, . . . are the amplitudes of the waves, to be deter- 
mined as functions of A. No cosine harmonies enter into this for- 
mula, because the m.m.f. distribution is symmetrical with respect 
to the center line 00^ of the exciting coil. To determine the ampli- 
tude of the nth harmonic An, multiply both sides of eq. (59) by 
sin nx d{nx), and integrate both sides between the limits x=0 and 

* For the general method of expanding a periodic function into a series of 
sines and cosines, see the author's Experimental Electrical Engineering, Vol. 2, 
pp. 222 to 227. 



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Chap. VII] M.M.F. OF DISTRIBUTED WINDINGS 125 

ic=7r. All the terms on the right-hand side vanish, except the one 
containing sin^ nx, and we have 

from which 

il„=4A/(n7r). ....... (60) 

Thus, the required series is 

h = 4A/7r (sin x + i sin 3a; + i sin 5a; + etc.) . . (61) 

This means that the amplitude of the fundamental wave is 4/7r 
times larger than the height h of the original rectangle; the ampli- 
tude of the third harmonic is equal to one-third of that of the fun- 
damental wave; the amplitude of the fifth harmonic is one-fifth of 
that of the fundamental wave, etc. In practical applications the 
fundamental wave is usually all we desire to follow, but in some 
special cases a few of the harmonics are important.^ 

Let now the winding of a phase be distributed in S slots per 
pole (Figs. 15 and 16), the distance between the adjacent slots 
being a electrical degrees. The conductors in every pair of slots 
distant by a pole pitch produce a rectangular distribution of the 
m.m.f . like the one shown in Fig. 34, or, what is the same, an equiv- 
alent series of sine-wave distributions. The m.m.fs. produced 
by the different coils are superimposed, and, since a sum of sine 
waves having equal bases is also a sine wave, the resultant m.m.f. 
also consists of a fimdamental sine wave and of higher harmonics. 
The fimdamental waves of the m.m.fs. of the several coils are dis- 
placed by an angle of a electrical degrees with respect to one 
another, so that the amplitude of the resultant wave is not quite 
S times larger than that of each component wave. The reduction 
coefficient, or the slot factor. A;,, is the same as that for the induced 
e.m.f. (Art. 28), because in both cases we have an addition of sine 
waves displaced by a electrical degrees, (see also prob. 20 in Art. 

1 This method of treating the mjn.fs. of distributed windings by resolv- 
ing the rectangular curve into its higher harmonics is due to A. Blondel. 
See his article entitled " Quelques propri^tfe gdn^rales des champs magn^ 
tiques toumants," VEdairage Electrique, Vol. 4 (1895), p. 248. Some authors 
consider the actual " stepped " curves of the m.m.f. or flux distribution, a 
procedure rather cumbersome, and in the end less accurate, in view of the 
fact that the higher harmonics are to a considerable extent wiped out by the 
currents in the rotor. 



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126 THE MAGNETIC CIRCUIT [Akt. 42 

28.) For the same reason, the value of the wmding-pitch factor, 
k^, deduced in Art. 29, holds for the m.m.fs. as well as for the 
induced e.m.fs. 

When adding the waves of the higher harmonics due to several 
coils, one must remember that an angle of a electrical degrees for 
the fundamental wave is equivalent to 3a electrical degrees for the 
third harmonic, 5a for the fifth harmonic, etc. Therefore, when 
using the formula (29) and Fig. 19, different values of a and of per 
cent pitch must be used for each harmonic, and in this connection 
the reader is advised to review Art. 30. In the practical problems 
given below the higher harmonics of the armature m.m.f. are dis- 
regarded altogether. The results so obtained are in a sufficient 
agreement with the results of experiments to warrant the great 
simplification so achieved. For the completeness of the treatment, 
and as an application of the general method, an analysis of the 
effect of the higher harmonics of an m.m.f. is given in Art. 45 
below. However, this article may be omitted, if desired, without 
impairing the continuity of the treatment in the rest of the book. 

Resolution of a Pulsating m.m.f. into Two Gliding m.m.fs. The 
reader is aware from elementary study that the pulsating m.m.f8. 
produced by two or three phases combine into one gliding (revolv- 
ing) m.m.f. in the air-gap. It is therefore convenient to consider 
even a single-phase pulsating m.m.f. as a combination of m.m.fs. 
gliding along the air-gap in opposite directions. In this wise, the 
m.m.fs. due to different phases are later combined in a simple 
manner. This method of treatment is similar to that used in 
mechanics, when an oscillatory motion is resolved into two rotary 
motions in opposite directions. Also in the analysis of polarized 
light a similar method of treatment is used. 

Take the first harmonic of the m.m.f. (Fig. 34) and assume the 
current in the exciting coil to vary with the time according to the 
sine law; then the amplitude of the m.m.f. wave also varies with 
the time according to the sine law. Imagine two m.m.f . waves, of 
half the maximum amplitude of the pulsating wave, gliding uni- 
formly along the air-gap in opposite directions; the superposition 
of these waves gives the original pulsating wave. One can see 
this by drawing such waves on two pieces of transparent paper and 
placing them in various positions over a sketch showing the pul- 
sating wave. It will be found that the sum of the corresponding 
ordinates of the revolving waves gives the ordinate of the pulsating 



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Chap. VII] M.M.P. OF DISTRIBUTED WINDINGS 127 

wave at the same point. Or else, represent the two gliding waves 
bj'^ two vectors of equal magnitude M, revolving in opposite direc- 
tions. The resultant vector is a pulsating one in a constant direc- 
tion, and varies harmonically between the values ±2Af. 

The analytical proof is as follows: Let the exciting current 
reach its maximum at the moment t=0. Then, if the amplitude of 
the m.m.f. wave at this instant is equal to A, the amplitude at any 
other instant t is equal to A cos 2;r/<. Therefore, the m.m.f. cor- 
responding to a point distant x from P and at a time t is equal to 
A cos 27:ft sin x. By a familiar trigonometrical transformation 
we have 

A sin X cos 27rft=iA sin (a:+2;r/0 +iA sin (x-2nft). (62) 

The right-hand side of this equation represents two sine waves, of 
the amplitude ^Ay gliding synchronously along the air-gap, that is, 
covering one pole pitch during each alternation of the current. 
The wave \A sin {x+2nft) glides to the left, because, with increas- 
ing t, the value of x must be reduced in order to get the same phase 
of the m.m.f . wave, that is, to keep the value of {x+2izf() constant. 
The other wave glides to the right, because, with increasing i, the 
value of x must be increased in order to obtain any constant value 
of (x—2nft), A similar resolution into two gliding waves can be 
made for each higher harmonic of the pulsating m.m.f. wave; the 
higher the order of a harmonic the lower the linear speed of its two 
gliding wave components. 

In practice it is usually required to know the relationship 
between the effective value i of the magnetizing current, the num- 
ber of turns n per pole per phase, and the crest value of one of the 
gliding m.m.f. waves. From the preceding explanation this rela- 
tionship for the fundamental wave is 

M = i(4A)fc6ni:\/2=0.9A:bm, .... (63) 

where M is the amplitude of each of the two gliding m.m.f s., m V2 
represents the maximum height h of the original rectangle, and the 
factor i is introduced because the amplitude of each gliding wave 
is one-half of that of the corresponding pulsating wave. The breadth 
factor fcft is the same as that used for the induced e.m.fs. (Arts. 27 to 
29) . Similar expressions can be written for each higher harmonic, 
remembering that their amplitudes decrease according to eq. (61), 



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128 THE MAGNETIC CIRCUIT [Art. 43 

and that a different value of ki must be used for each harmonic. 
The value of ikf is calculated so as to produce the required revolv- 
ing flux, as is explained in Chapters IV, V, and VI. From eq. (63) 
either n or i, or their product can be determined. 

Prob. 1. A single-phase four-pole induction motor has 24 stator slots, 
two-thirds of which are occupied by the winding; there are 18 con- 
ductors per slot. The average reluctance of the active layer is 0.09 rel. 
per square centimeter. What current is necessary to produce a pulsating 
flux of such a value that the maximum flux density due to the first 
harmonic is 5 kl./sq.cm., when the secondary circuit is open? 

Ans. 8.3 amp. 

Prob. 2. Show that in the preceding problem the difference between 
the actual flux per pole and its fundamental is less than 2 per cent. 

Prob. 3. Show that, if in Fig. 34 the angle x is counted from the 
crest of the first harmonic, the expansion into the Fourier series is similar 
to eq. (61), except that cosines take place of the sines, and the terms 
are alternately positive and negative. 

43. The M.M.F. of Polyphase Windings. Ck)nsider a two- 
phase winding of the stator of an induction motor (Fig. 35a) ; let 

,2 Armature T 



m. 



llt i m ^ 



P ^ -- ^ Slots* -* ^^ 
Fig. 35a. — ^A two-phase winding. 

the current in phase 1 lead that in phase 2 by JT, or by 90 electrical 
degrees. A little reflection will show that the resultant m.m.f. of 
the two phases glides from right to left : Let the current in phase 1 
reach its maximum at the instant t=0; at this instant the current 
in the coil 2 is zero, and the m.m.f. wave is distributed imiformly 
imder the coil 1 ; at the instant t=iT the current in phase 1 is zero, 
and the m.m.f. is distributed under the coil 2. At intermediate 
instants both coils contribute to the resultant m.m.f., so that its 
maximum occupies a position intermediate between the centers 
Oi and O2 of the coils 1 and 2. 

The actual rectangular distribution of the m.m.f. due to each 
phase can be replaced by a fimdamental sinusoidal one and its 
higher harmonics, as in Fig. 34:. The pulsating fimdamental m.m.f. 
of each phase can be replaced by two waves of half the ampli- 
tude, gliding synchronously in opposite directions. Let the wave 



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Chap. VII] M.M.F. OF DISTRIBUTED WINDINGS 129 

due to phase 1, and gliding to the left, be denoted by Li, and that 
due to phase 2 by L2. Let the corresponding waves gliding to the 
right be denoted by Ri and R2. Disregarding the higher har- 
monics, the resultant m.m.f . is due to the combined action of the 
four gliding waves Li, L2, Ri and R2. At the instant i=0 the 
crest of the wave Li is at the point Oi; at the instant t=iT the 
crest of the wave L2 is at the center O2 of the coil 2. Consequently, 
at the instant t=0 the crest of the wave L2 is 90 electrical degrees 
to the right of O2, or it is at Oi. Thus, the waves Li and L2 actu- 
ally coincide in space, and form one wave of double the amplitude. 
The crest of the wave Ri is at the point Oi when t=0; the crest 
of R2 is at the point O2 when t=iT. Therefore, at i=0 the crest 
of R2 is 90 electrical degrees to the left of the point O2, and the 
waves Ri and R2 travel at a distance of 180 electrical degrees from 
each other. But two such waves cancel each other at all points 
and at all moments, so that there is no resultant R wave. Thus 
the resultant fxmdamental wave of m.m.f. in a two-phase machine 
is gliding. Its amplitude is twice as large as that of either of the 
component gliding m.m.f s. of the two phases, which components 

,8 Armature .2 



ZL , |r^ 



fsi l I ^J T l 66 l 1 66 ] [ it ] ( g3 ] [ 5 ^ ^ fj 

>- ^SlotB^ -^- 

R L 

Fig. 356. — ^A three-phase winding. 

are expressed by eq. (63). If the current in phase 2 were leading 
with respect to that in phase 1, the L fluxes would cancel each 
other and the resultant flux would travel from left to right. 

Consider now a three-phase winding (Fig. 356) and call the 
m.m.fs. which glide to the left, and which are due to the separate 
phases, by Li, L2, and L3 respectively. Let the waves which 
travel to the right be denoted by Ri, R2, and R3. Assume the cur- 
rent in phase 2 to be lagging by 120 electrical degrees, or by JT", 
with respect to that in phase 1, and the current in phase 3 to be 
lagging by ^T with respect to that in phase 2. By a reasoning 
similar to that given for the two-phase winding above it can be 
shown that the three L waves coincide in their position in space, 
and give one gliding wave of three times the amplitude of each 
wave. The three R waves are relatively displaced by 240 elec- 



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130 THE MAGNETIC CIRCUIT [Abt. 43 

trical degrees, or, what is the same, by 120 electrical degrees; 
hence, their m.mis. mutually cancel at each point along the 
air-gap. This can, be proved by drawing three sine waves dis- 
placed by 120 degrees and adding their ordinates point by point; 
or else one can replace each wave by a vector, and show that the 
sum of the three vectors is zero because they form an equilateral 
triangle. 

The reasoning given for the two- and three-phase windings can 
be extended to any number of symmetrical phases, say m, pro- 
vided that the windings are displaced in space by 360/m electrical 
degrees, and also provided that the currents in these windings are 
displaced in time by 1/mth of a cycle. The gliding fundamental 
waves due to each phase which go in one direction are in phase 
with each other, and, when added, give a wave m times larger 
than that expressed by eq. (63) ; while the fimdamental waves 
going in the opposite direction are displaced in space by 720/m 
electrical degrees, and their combined m.m.f. is zero. The direc- 
tion in which the resultant m.m.f. travels is from the leading to 
the lagging phases of the winding. Thus, for any synmietrical 
m-phase winding 

Af=0.9A;6?nm, (64) 

wliere M denotes the amplitude of the fimdamental sine wave of 
the resultant gliding m.m.f ., n is the number of turns per pole per 
phase, and i is the effective value of the ciurent in each phase. 

Prob. 4. It is desired to build a 60 horse-power, 550-volt, 4-pole, 
Y-connected induction motor, using a stator punching with 4 slots per 
pole per phase, and a winding pitch of one hundred per cent. The required 
maximum m.m.f. per pole is estimated at 1550 ampere-tums. What is 
the total required number of stator turns (for all the phases) if the mag- 
netizing current must not exceed 25 per cent of the full-load current? 
The estimated full-load efficiency is 92 per cent, the power factor at 
full load is about 90 per cent? Ans. Not less than 504. 

Prob. 5. What is the required number of turns in the preceding 
problem, if the stator winding is to be delta-connected and to have a 
winding pitch of about 75 per cent? Ans. Not less than 936. 

Prob. 6. What is the amplitude of the first harmonic of the total 
armaturere action in a 100-kva, 440-volt, 6-pole, two-phase alternator 
with non-salient poles? The stator has 72 slots; the coils lie in slots 1 
and 9; the number of conductors per slot is C^. In practice, the arma- 
ture reaction must not exceed a certain limit, and this helps to determine 
the permissible value of C* Ans. 4800Ca amp.-tums. 



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Chap. VII] M.M.F. OF DISTRIBUTED WINDINGS 131 

Prob. 7. Plot the actual " stepped " curves of m.m.f. distribution 
for a two-phase winding with three slots per pole per phase, for the fol- 
lowing instants: 1 = 0, t=^-^T, and t = iT, Compare the maximum and 
the average m.m.f. of the actual distribution with those of the first 
harmonic* 

Prob. 8. Solve the preceding problem foi- a three-phase winding 
with 2 slots per pole per phase, and a winding pitch of f . Take two 
instants, <=0 and <=tV^> ^^ show that for the instants ^iV^» A^> A^> 
etc., the m.m.f. distribution is the same as for <=0, while for ^T, ^^T, 
etc., the m.m.f. distribution is the same as for t = ^T, 

Prob. 9. Prove directly that two equal pulsating sine waves of m.m.f . 
or flux, displaced by 90 electrical degrees in space and in time relatively 
to each other, give a gliding sine wave, the amplitude of which is equal 
to that of each pulsating wave. Solution: The left-hand side of eq. 
(62) gives the value of the m.m.f. at a point x and at an instant t, due 
to phase 1 ; the m.m.f . produced at the same point and at the same instant 
by the phase 2mAmi{x+in) cos {2nft—in), Adding the two expressions 
gives A sin {x +2nft), which is a left-going wave of amplitude A. 

Prob. 10. Prove, as in the preceding problem, that the three pulsating, 
sine waves of m.m.f. produced by a three-phase winding, give together 
a gliding m.m.f., the amplitude of which is 50 per cent larger than that 
of each pulsating wave. 

Prob. 11. Prove by the method given in problem 9 above that m 
pulsating m.m.f. waves displaced in space and in time by an electrical 
angle 27r/m produce a gliding m.m.f. the amplitude of which is im times 
larger than that of each pulsating wave. See Arnold, Wechselstrom- 
technik, Vol. 3 (1908) p. 302. 

44. The M.M.FS. in a Loaded Induction Machine.^ Eq. (64) 
gives the liiagnetizing current t of an induction motor at no-load, 
i.e., when the rotor is running at practically synchronous speed, so 
that the secondary currents are negligible. When the motor is 
loaded, the useful flux which crosses the air-gap is due to the com- 
bined action of the primary and the secondary currents. In com- 
mercial motors the flux at full load is but a few per cent below that 
at no load, the difference being due to the impedance drop in the 

* Problems 7 and 8 are intended to acquaint the student with the usual 
method of calculation of the m.m.fs. of distributed windings and to show the 
advantage of Blondel's method used in the text. For numerous stepped 
curves and calculations, see Boy de la Tour, The Induction Motor, Chapter IV. 

'The treatment in this article presupposes a general knowledge of the 
equivalent performance diagram of induction machines; the purpose of 
the article being to deduce the exact numerical relations. This article and 
the one following can be omitted without impairing the continuity of treat- 
ment in the rest of the text. 



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132 THE MAGNETIC CIRCUIT [Art. 44 

primary winding, the same as in a transformer. Therefore, the 
net number of exciting ampere-turns, Af, is approximately the 
same as at no load. This means that the geometric sum of 
the m.m.f s. produced by the primary, and the secondary currents at 
any load is nearly equal to the m.m.f. due to the primary winding 
alone at no load. In this respect the induction motor is similar to 
a transformer. 

(a) CofctiZa^ion o/^yie Secondary Cwrren^. Knowing the primary 
full-load current, thfe secondary full-load current can be calcu- 
lated from the required coxmter-m.m.f.; the procedure can be best 
illustrated by an example. In the motor given in prob. 4 above, 
the full-load current is estimated at 57 amp.; taking the direction 
of the vector of the applied voltage as the axis of reference, the 
full-load current can be represented as 51.3— ^24.8 amp. The 
magnetizing current, 0.25X57= 14.25, is practically in quadrature 
with the applied voltage, because it is in quadrature with the 
induced counter e.m.f., the same as in a transformer. The full- 
load current of 57 amp. contains a component which supplies the 
iron loss in the stator; we estimate it to be equal to about 1.1 amp. 
(2 per cent of the input). Thus, the component of the primary 
current, the action of which must be compensated by the second- 
ary currents, is (51.3 -/24.8) - (1.1 -7*14) = 50.2 -7IO.8 amp., or its 
absolute value is 5.14 amp. This is called the current transmitted 
into the secondary , or the secondary current reduced to the 'primary 
circuit. This current produces a maximum m.m.f . of 0.9 X 3 X 0.958 
X 42X51.4 =5580 amp.-tums. 

Let the rotor be provided with a three-phase winding, with 5 
slots per pole per phase, and let the winding pitch be 13/15. The 
number of slots is selected so as to be different from that in the sta- 
tor, in order to insure a more uniform torque, and to reduce the 
fluctuations in the reluctance of the active layer. We have, 
according to eq. (64), that 5580 = 0.9 X3 X0.935 X (m), from which 
m=2210 amp.-tums. Certain practical considerations, for 
instance, the value of the induced secondary voltage, usually 
limit the choice of one of these factors; then the other factor 
also becomes definite. If, for instance, the rotor is to have 10 
conductors per slot, the secondary current will be about 89 amp. 
The secondary i?r loss is determined by the desired per cent slip; 
knowing the secondary current and the number of turns, the 
necessary size of the conductor can easily be calculated. 



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Chap. VII] M.M.F. OF DISTRIBUTED WINDINGS 133 

Sometimes the secondary winding consists of coils individuaUy 
short drcwited; this is an intermediate type of winding between 
an ordinary squirrel-cage winding and a three-phase winding 
such as is used with slip-rings. Let the foregoing motor be 
provided with such a winding, of the two-layer type, and let the 
rotor have 71 slots, 6 conductors per slot, the coils being placed 
in slots 1 and 14. In formula (64) m stands for the number of 
symmetrically distributed phases, the current in each phase being 
displaced in time by 27r/m with respect to that in the next phase. 
In the winding under consideration, each coil represents a phase, 
and one has to go over a pair of poles until one finds the next coil 
with the current in the same phase. Thus, in this case, the num- 
ber of secondary phases is equal to the number of slots per pair of 
poles, or m= 35.5. Each coil has 3 turns, but there is oidy one coil 
per pair of poles, so that n= 1.5. Substituting these values into 
eq. (64), and also M=5580, ^6=0.912, we find i= 128 amp. As a 
matter of fact, in this case it is not necessary to decide what the 
values of n and m are, because eq. (64) contains only the product 
mn, which is the total number of turns per pole. Thus, in our case 
mn={7lXS)/4:. 

Formula (64) holds also for a squirrel-cage winding, the number 
of secondary phases being equal to the number of bars per pair of 
poles. Since there is but one bar per phase, each bar can be con- 
sidered as one-half of a turn, and in formula (64) n=0.5 and ki,= 1, 
so that it becomes 

M=0.45iC2/p, (64a) 

where C2 is the total number of rotor bars, and p is the number of 
poles. Or else, one may say that the total number of turns per 
pole is equal to one-half the number of bars per pole, so that 
mn=iC2/p. This again gives eq. (64a). For a direct proof of 
formula (64a) see problem 15 below. Applying this formula to 
the same rotor with 71 slots we find that the current per bar is 
700 amp. 

(6) The Equivolent Secondary Winding Reduced to the Primary 
Circuit. When investigating the general theory of the induction 
motor or calculating the characteristics of a given motor, it is con- 
venient to replace the actual rotor winding by an equivalent wind- 
ing identical with the primary winding of the motor. In this case 
the primary current transmitted into the secondary is equal to the 



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134 THE MAGNETIC CIRCUIT [Akt. 44 

actual secondary current (one to one ratio of transformation), and 
the primary and the secondary voltages induced by the useful flux 
are also equal. Each electric circuit of the stator then can be 
combined with the corresponding rotor circuit. In this manner 
the so-called '' equivalent diagram " of the induction motor is 
obtained,^ a way of representation which greatly simplifies the 
theory of the machine. 

Let i2 be the secondary current in the coils or bars of the actual 
rotor, and i'2 that in the equivalent rotor. The coimter-m.m.f. 
of both rotors must be the same, this being the condition of their 
equivalence, so that 

0.9A;52W2W2t2 = 0,9kiiminii2', 
from which 

i27i2=(rn2/mi)(kb2n2/hini) (65) 

This is the ratio of current transformation in an induction motor. 
The ratio of transformation of the voltages is different, namely, 

e2'/e2 = hini/kb2n2 (66) 

In an ordinary transformer e2Ve2= 1*2/^2' = ^1/^2, because there 
Afei = ^62=l; and m2 = mi = l. For this reason, the induction 
motor is sometimes regarded as a generalized transformer. 

Taking the product mie for the actual and the equivalent rotor 
it will be found that the total electric power input is the same in 
both, provided that the same phase displacement is preserved in 
the equivalent rotor as in the original one. The latter condition is 
essential in order that the operating characteristics of the two 
machines be the same. This means (a) that the total i^r loss of 
the equivalent rotor must be equal to that of the original rotor, in 
order to preserve the same slip, and (6) that the leakage react- 
ances of the two rotors must affect the power factor of the 
primary current in the same way. 

Let r2 and r2' be the resistances of the actual and of the equiva- 
lent rotor, per pole per phase. We have the condition that 

m2i2^2 = 'fnii2^2 (67) 

1 Chas. P. Steinmetz, AUernating Current Phenomena (1908), p. 249; 
Elements of Electrical Engineering (1905), p. 263. 



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Chap. VII] M.M.F. OF DISTRIBUTED WINDINGS 135 

Substituting the ratio of 127^2 from eq. (65) we find 

r2/r2=^(jni/m2){h\n\/h2n2)? .... (68) 

For a transformer this equation reduces to the familiar expression 
r2'A2=(ni/n2)2.i 

The ratio of the inductances is the same as that of the resist- 
ances; this can be proved as follows: In order that the equivalent 
winding may have the same effect on the power factor of the motor 
as the actual winding, the equivalent winding must draw from the 
line an equal amount of reactive volt-amperes, due to its leakage 
inductance. The magnetic energies stored in the two rotor wind- 
ings must therefore be equal, and we have, according to eq. (104) 
in Art. 58, 

m2'ii2^I/2 = mi.iiV^L'2, (69) 

where L2 and L'2 are the leakage inductances of the real and the 
equivalent rotor windings, per pole per phase. The form of this 
equation is the same as that of eq. (67) ; therefore, substituting 
again the ratio of i'2A2 from eq. (65) an expression is obtained for 
the ratio of U2/L12 identical with that given by eq. (68), namely 

L2/L2^{mi/m2){hi^\/h2n2)^ (70) 

This result could also be foreseen from the fact that the reactances 
and the resistances enter symmetrically in the equivalent diagram, 
and relation (68) holds therefore for the reactances X2 and X2\ 
But in the equivalent diagram the secondary and the primary fre- 
quency is the same, so that the ratio of the inductances is equal to 
that of the reactances; this gives eq. (70). 

It must be clearly imderstood that the expressions (68) and (70) 
refer to the resistances and inductances per pole per phase. When 
the windings of a phase are all in series, both in the stator and in the 
rotor, the same ratio holds of course for the resistances per phase; 
otherwise the actual connections must be taken into consideration, 
keeping in mind that the total i^r loss must be the same in the 
equivalent winding as in the actual one. Having obtained the re- 
sistance of the equivalent winding per pole, the turns are connected 
in the same way as the stator turns. This fact must be remembered 
in particular when dealing with individually short-circuited coils 

* See the author's Experimental Electrical Engineering, Vol. 2, p. 77. 



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136 THE MAGNETIC CIRCUIT [Aet. 45 

in the rotor, or with a squirrel-cage winding. In these two cases 
the individual coils or bars in the rotor are all in parallel, while the 
stator coils of a phase are usually all in series, or in two parallel 
groups. In the case of a squirrel-cage winding the resistance r2 
includes that of a bar, of two contacts with the end-rings, and of 
the equivalent resistance of a section of the two end-rings.^ 

Prob. 12. In a 300 horse-power, Y-connected, 14-pole induction 
motor the full-load current is estimated to be 310 amp. The primary 
winding consists of 336 turns placed in 168 slots; the winding-pitch is 
0.75. What is the minimum number of bars in the squirrel-cage second- 
ary winding, if the current per bar must not exceed 800 amp.? The 
secondary counter-m.m.f. is equal to about 90 per cent of the primary 
m.m.f. Ans. 208. 

Prob. 13. What must be the resistance of each secondary bar in the 
preceding problem (including the equivalent resistance of the adjoining 
segments of the end-rings and also of the contacts) if the slip at full load 
is to be about 4 per cent.? Hint: The per cent slip is equal to the iV 
loss in the rotor, expressed in per cent of the power input into the second- 
ary. If a; is the iV loss in the rotor, expressed in horse-power, we have 
that X = 0.04 (300 + x) . Ans. 70 microhms. 

Prob. 14. The motor with the individually short circuited second- 
ary coils, that is used as an illustration in the text above, is to be 
investigated with respect to its performance. By what factor must the 
actual resistance and inductance of each secondary coil be multiplied 
in order to obtain the equivalent resistance and inductance per primary 
phase? Also by what factor must the equivalent current be multiplied 
in order to obtain the actual current in each secondary coil? 

Ans. 297; 0.402. 

Prob. 15. Prove formula (64a) directly, by considering the m.m.fs. 
of the individual bars. Solution : At any instant the currents in the bars 
under a pole are distributed in space according to the sine law, because 
the gliding flux which induces these currents is sinusoidal. The average 
current per bar is fV2^X(2A) =0.9t. The number of turns per pole is 
C2/2P, and all these turns are active at the crest of the m.m.f. wave. 
Therefore, M ^0.9i(C2/2p). 

45. The Higher Harmonics of the M.M.FS. In the preceding 
study, the effect of the higher harmonics in the m.m.f . wave was dis- 
regarded. In fact, these harmonics usually exert a negligible 
influence upon the operation of a good polyphase induction motor, 
under normal conditions. These m.m.f. harmonics move at lower 
speeds than the f xmdamental field ; therefore, the fluxes which they 

* See the author's Electric Circuit; also E. Arnold, Die Wechselstromtechnik, 
Vol. 5, Part I (1909), p. 57. 



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Chap. VII] M.M.F. OF DISTRIBUTED WINDINGS 137 

produce cut the secondary conductors at comparatively high rela- 
tive speeds; thus, secondary currents are induced which wipe out 
these harmonics to a considerable degree. There are practical cases, 
however, in which some one particular harmonic becomes of some 
importance, and affects the operation of the machine, particulsCrly 
at starting. For this reason the following general outline of the 
properties of the higher harmonics in the m.m.f . is given.^ 

In a single-phase machine (Fig. 34) all the higher harmonics of 
the m.m.f . are pulsating at the same frequency as the fimdamental 
wave, but the width of the nth harmonic is only 1/nth of that of 
the fimdamental wave. Each pulsating harmonic can be replaced 
by two gliding harmonics of half the amplitude, one left-going, the . 
other right-going. The linear velocity of these gliding m.m.fs. is 
only 1/nth of that of the fundamental gliding waves, because they 
cover in the time ^T a distance equal only to their own base, PQIn 
(180 electrical degrees) . With one slot per pole, the amplitudes of 
the higher harmonics decrease according to eq. (61), but with more 
than one slot, or with a fractional-pitch winding they decrease 
more rapidly, because different values of A^ must be taken for each 
harmonic (see Art. 30 above). 

In a two-phase machine, consider (Fig. 35a) the gliding waves 
L„ and R^, of the nth harmonic. For this harmonic, the distance 
between 0\ and O2 is equal to \Tzn electrical degrees. At the 
instant i=0 the crest of the wave L^^ is at the point 0\\ at the 
instant i^\T the crest of the wave L^g is at the point O2. There- 
fore, the two waves travel at a relative distance of J;r(n — 1) elec- 
trical degrees, considering the base of the nth harmonic as equal 
to its own 180 electrical degrees. In a similar manner, the distance 
between the crests of the two right-going waves is found to be 
equal to J;r(n4- 1) electrical degrees. We thus obtain the following 
table of the angular distances between the waves due to the two 
phases: 

Order of the harmonic 

Distance between the two Ln waves 
Distance between the two Rn waves 

The waves which travel at a distance 0, 2;r, 4;r, etc., are simply 
added together, while those at a distance tt, 3;r, 5?:, etc., cancel each 

1 For a more detailed treatment see Arnold, Wechselstromtechnikf Vol. 
3 (1904), Chapter 13, and Vol. 5, part I (1909), Chapter 9. 



1 


3 


5 


7 


9 


11 


13 





It 


2n 


St: 


47r 


5;r 


6;r 


7: 


In 


Sn 


4;r 


dn 


6n 


Ttt 



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138 THE MAGNETIC CIRCUIT [Akt. 45 

other. Thus, in a two-phase machine, the 3d, 7th, 11th, etc., 
harmonics travel against the direction of the main m.m.f ., while the 
5th, 9th, 13th, etc., harmonics travel in the same direction as the 
fundamental m.m.f., though at lower peripheral speeds. 

Applying a similar reasoning to a three-phase winding (Fig. 
356) we find that the three L„ waves travel at a relative distance of 
|;r(n — 1), while the relative distance between the three R„ waves is 
i7t(n+l) electrical degrees. We thus obtain the following table of 
the angular distances between the waves due to the three phases: 

Order of the hannonic 1 3 6 7 9 11 13 15 

Distancebetween the three Ln waves... |« {jt V^ V^ ^ 
Distance between the three Rn waves. .. in in V^ V«: ^^ 

The component waves, of any harmonic, which travel at a distance 
zero from each other, are simply added together, and give a resul- 
tant wave of three times the amplitude of the component. The 
three waves which travel at an angular distance of ^tt or one of 
its multiples from each other give a sum equal to zero. Thus, in a 
three-phase machine, the 1st, 7th, 13th, etc., harmonics travel in 
one direction, while the 5th, 11th, 17th, etc., harmonics travel 
against the direction of the fimdamental m.m.f. The higher the 
order of a harmonic the lower its peripheral speed. The harmonics 
of the order 3, 9, 15, etc., are entirely absent. 

Prob. 16. What are the amplitudes of the fifth and the seventh 
harmonics, in percentage of that of the fundamental wave, for a three- 
phase winding placed in 2 slots per pole per phase, when the winding- 
pitch is 5/6? Ans. 1.4 and 1.0 per cent respectively. 

Prob. 17. Show that, in order to eliminate the nth hannonic in 
the m.m.f. wave, the winding-pitch must satisfy this condition; namely, 
r/7:^{2q + l)/nj where r is defined in Fig. 16, and q is equal to either 0, 
1, 2, 3, etc. Hint: Cos ir^ must be =0. 

Prob. 18. Investigate the direction of motion of the various har- 
monics of the m.m.f. in a sjmimetrical m-phase system. 

Prob. 19. Show that only the nth harmonic in the m.m.f. wave, 
due to the nth harmonic in the exciting current, moves synchronously 
with the fundamental gliding m.m.f., and therefore distorts it perma- 
nently. 

Ptob. 20. A poorly designed 2-phase, 60-cycle induction motor has 
4 poles, 1 slot per phase per pole, and a winding pitch of 100 per cent. 
At what subnsynchronous speed is it most likely to stick? Hint: The 
torque due to any harmonic reverses as the motor passes through the 
corresponding sub-sjrnchronous speed. Ans. 360 r.p,m. 



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CHAPTER VIII 

ARMATURE REACTION IN SYNCHRONOUS 
MACHINES 

46. Armature Reaction and Armature Reactance in a Syn- 
chronous Machine. When a synchronous machine carries a load, 
either as a generator or as a motor, the armature currents, being 
sources of m.m.f ., modify the flux created by the field coils, and 
thus influence the petf ormance of the machine. Fig. 36 shows an 




« * Direction of Botation 

Fig. 36 — The flux distribution in a single-phase synchronous machine 

under load. 



instantaneous flux distribution in the simplest case of a single- 
phase alternator, with one slot per pole; the armature conductors 
are marked a and b. With the directions of the armature and field 
currents indicated in the sketch, the flux is crowded toward the 
right-hand tips of the poles. In order to show this, imagine two 
fictitious conductors a' and V with currents equal and opposite 

139 



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140 THE MAGNETIC CIRCUIT IArt. 46 

to those in the actual conductors a and b respectively. The addi- 
tion of these fictitious conductors does not modify the armature 
m.mi. because they neutralize each other. The conductor a' 
may be considered as forming a turn with a, while 6' forms a turn 
with 6. It will be seen that the m.m.f . of the coil aa' assists that 
of the field coil A, while the m.m.f. of the coil bV is opposite to that 
of the field coil B. 

The armature current in the coil ab not only distorts the no- 
load field, but also reduces the total flux per pole. This may be 
seen by considering the flux in the four parts of the air-gap, marked 
X, y, x\ and r/', where x=a:' and y=y'. The sum of the fluxes in 
the portions y and y' is the same as without the armature current, 
because the flux density in the part y is increased by the same 
amount by which it is reduced in the part y' (neglecting satura- 
tion). But in the parts x and x' the flux is reduced by the arma- 
ture m.m.f ., so that the total result over the pole-pitch is a reduc- 
tion in the value of the no-load flux. The position of the armature 
conductors and the direction of the armature currents have been 
selected arbitrarily. They can be chosen so that the flux will be 
crowded toward the left-hand tips of the poles, or so that the total 
flux will be increased by the armature m.m.f., instead of being 
reduced. The influence of the armature currents, in modifying the 
value of the field flux and distorting it, is called the armature reac- 
tion. The armature reaction is measured in ampere-turns, since 
it is a magnetomotive force. 

In addition to the general distortion of the field by the arma- 
ture currents, there is a local distortion aroimd each armature 
conductor. This distortion does not extend into the pole shoes, 
but is limited to the slots and the air-gap; it is indicated in Fig. 36 
by ripples in the flux around a and b. These ripples may be 
regarded as a result of the superposition upon the main flux of the 
local fluxes <Po and (Pj, excited by the armature currents. While 
these local fluxes, shown by the dotted lines, have no real existence, 
except around the end connections of the armature conductors, 
it is convenient to consider them separately. They are purely 
alternating fluxes, in phase with the currents with which they are 
linked, so that they induce in the armature windings alternating 
e.m.fs. in a lagging phase quadrature with the c\irrents. 

The effect of these local fluxes upon the voltage of the machine 
is represented by a certain armature reactance, because the effect is 



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Chap. VUI] REACTION IN SYNCHRONOUS MACHINES 141 

the same as if the armature winding created no leakage fluxes 
around it, but a separate reactance coil were connected in series 
with each armature lead. The calculation of the armature react- 
ance, or of the local fluxes, is treated in Art. 67, the subject of this 
chapter being armature reaction only, that is, the effect of the load 
upon the main magnetic circuit. In the numerical problems of this 
chapter, for the solution of which it is necessary to know the value 
of the armature reactance, this value is given. It is not quite 
correct, strictly speaking, to separate the local distortion of the 
main flux as a phenomenon by itself; moreover, the separation is 
somewhat indefinite and arbitrary. However, the flux so separated 
is comparatively small, and the treatment of the armature reaction 
proper is thereby greatly simplified. 

The distribution shown in Fig. 36 varies from instant to instant 
because the relative position of the armature changes with refer- 
ence to the poles, as well as the value of the armature current. 
Besides, there are usually two or three armature phases, and sev- 
eral slots per pole per phase. It would be out of the question to 
calculate the actual fluxes for each instant and to take into accoimt 
their true influence upon the e.m.f . induced in the armature. In 
practice, certain approximate average values of armature reaction 
and of armature reactance are employed, which permit one to 
predict the actual performance of a machine with a sufficient 
accuracy. 

In the case of a synchronous generator (alternator) the problem 
usually presents itself in the following form: It is required to pre- 
determine the field ampere-turns necessary for a prescribed ter- 
minal voltage at a given load. Knowing the resistance and the 
leakage reactance of the armature, the voltage drop in the arma- 
ture is added geometrically to the terminal voltage; this gives the 
induced voltage in the machine. Knowing from the no-load satura- 
tion curve the required net excitation at this voltage, and correct- 
ing it for the effect of the armature reaction, the necessary field 
ampere-turns are obtained. The results of such calculations for 
different values of the armature current and for various power 
factors, plotted as curves, are called the load characteristics of the 
alternator. 

In the case of a synchronous motor the terminal voltage is usu- 
ally given, and it is required to determine the field excitation such 
that, at a given mechanical output, the input to the armature be at 



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142 THE MAGNETIC CIRCUIT [Abt. 46 

a given power-factor; a leading power-factor is usually prescribed, 
in order to raise the lagging power-factor of the whole plant. The 
problem is solved in like manner to that of the generator, by 
taking into account the proper signs when calculating the reactance 
drop and the armature reaction. The results, plotted in the form 
of curves, are called the phase characteristics, or V-curves of a 
sjrnchronous motor.^ 

It will be seen from Fig. 36 that the crowding of the flux to one 
pole-tip, by the armature currents, is primarily due to the fact that 
the poles shown there are projecting or salient, so that the reluc- 
tance along the air-gap is variable. With non-salient poles the 
flux is simply shifted side wise without being distorted. Therefore, 
before going into the details of the calculation of armature reaction 
in machines with salient poles we shall first consider (in the next 
article) the case of a machine with non-salient poles. 

Prob. 1. Draw the distribution of the flux, similar to that shown 
in Fig. 36, when the armature conductors are opposite the centers of the 
poles, and when they are somewhere between the adjacent pole-tips. 

Prob. 2. Explain the details of the flux distribution in Fig. 36, by 
means of a hydraulic analogy, assuming A and B to represent two main 
centrifugal piunps, and a and 6 to be two smaller pumps placed in the 
stream. 

Prob. 3. Let each field cofl in Fig. 36 have N timis, and let the 
exciting current be /; let the number of conductors at a be C„ and the 
instantaneous value of the armature current i. What is the total flux 
per pole, if the average permeance of the machine per pole is (P perms 
per electrical radian, and the angles and a; are in electrical radians? 

Ans. (2NId-Cgix)(P. in maxwells. 

Prob. 4. Let a synchronous machine be loaded in such a way that 
the armature current reaches its maximum when the conductors a and b 
(Fig. 36) are opposite the centers of the poles, in other words, the current 
is in phase with the e.m.f. which would be induced at no load. Prove 
that (neglecting saturation) the average flux per pole during a complete 
cycle is the same as without the armature reaction, but is crowded to 
the leading tip of the pole, i.e., in the direction of rotation in the case of 
a motor, and to the trailing tip, or against the direction of rotation when 
the machine is working as a generator. Hint: The flux is weakened as 
much in the position x of the conductors as it is strengthened in the 
S3nnmetrical position x'; the distortion is in the same direction in both 
positions. 

* See the author's " ExperimerUal Electrical Engineering" Vol. 2, p. 121; 
also his "Essays on Synchronous Machinery," Oeneral Electric Review^ 
1911, p. 214. 



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Chap. VIIl] REACTION IN SYNCHRONOUS MACHINES 143 

Prob. 6. Let a synchronous machine be loaded in such a way that 
the armature current reaches its maximum when the conductors a and h 
(Fig. 36) are midway between the poles, in other words, when the current 
is displaced by 90 electrical degrees with respect to the e.m.f. induced 
at no load. Prove that the average distortion during a complete cycle 
is zero, but that the flux is weakened if the armature current lags behind 
the induced e.m.f., and is strengthened by a leading current. Hint: 
The flux is weakened in both of the symmetrical positions, x and x', of 
the conductors, but the distortion is in opposite directions. 

Prob. 6. In a single-phase synchronous machine the armature 
current reaches its maximum when the armature conductors are dis- 
placed by an angle ^ with respect to the centers of the poles;* prove that 
the field is distorted by the component i cos </f of the current and is 
weakened or strengthened by the component i sin i/f, 

47. The Performance Diagram of a Synchronous Machine with 
Non-Salient Poles. Let, in a machine with non-salient poles, the 
field winding be placed in several slots per pole, so that the field 
m.m.f . in the active layer of the machine is approximately distrib- 
uted according to the sine law. Consider the machine to be a 
polyphase generator supplying a partly inductive load. The ampli- 
tude of the first harmonic of the armature reaction has the value 
given by eq. (64) in Art. 43, and revolves synchronously with the 
field m.m.f ., as is explained there. Since the sum of two sine waves 
is also a sine wave, the resultant m.m.f. is also distributed in the 
active layer of the machine according to the sine law. 

To deduce the phase displacement, in space, between the two 
sine waves, consider the coil a b (Fig. 36) to be one of the phases of 
the polyphase armature winding. For reasons of symmetry, the 
maximum m.m.f. produced by a polyphase winding is at the center 
of the coil in which at that particMlar moment the current is at a 
maximum. Assume first that the current in the phase a 6 reaches its 
maximum when the conductors a and 6 are opposite the centers of 
the poles. The maximum armature m.m.f. at that instant is dis- 
placed by 90 electrical degrees with respect to the center lines 
of the poles. The direction of the armature current is determined 
by the well-known rule, and it is found to be such that the arma- 
ture m.m.f . lags behind that of the pole, considering the direction 
of rotation of the poles as positive. Since both m.m.fs. revolve 
synchronously, this angle between the two m.m.f. crests is pre- 

* The angle <p is different from the external phase-angle between the 
current and the terminal voltage; see Fig. 37. 



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144 THE MAGNETIC CIRCUIT [Abt. 47 

served all the time. Thus, in a polyphase generator, the armature 
rn.mi. lags behind the field m.m.f. by 90 electrical degrees in 
space, when the currents are in phase with the voltages induced at 
no-load. This statement is in accord with that in problem 4 in the 
preceding article, because, if each phase shifts the flux against the 
direction of rotation, all the phases together simply increase the 
result. 

Let now the currents in the armatiu^ windings be lagging 
90 electrical degrees behind the corresponding e.m.fs. induced 
at no-load. This simply means that the armature m.m.f . is shifted 
further back by 90 degrees as compared to the case considered 
before; therefore, the angle between the field rn.mi. and the 
armature m.m.f. is 180 electrical degrees, and the two m.m.fs. are 
simply in phase opposition. This is in accord with the statement 
in prob. 5. 

From the two preceding cases it follows that, when in a syn- 
chronous machine with non-salient poles the currents lag by an 
angle electrical degrees (Figs. 37 and 38) with respect to the 
induced voltage at no-load, the armatiu^ m.m.f . wave lags by an 
angle of 90 + ^ electrical degrees behind the field m.m.f. wave. In 
the case of a generator with leading currents the angle ^ is negative ; 
in a synchronous motor is larger than 90 degrees. 

Let, in Fig. 37, i be the vector of the current in one of the phases, 
and let e be the corresponding terminal voltage, the phase angle 
between the two being ^. Adding to 6 in the usual way the ohmic 
drop ir in the armature, in phase with i, and the reactive drop ix 
in leading quadrature with t, the induced voltage E in the same 
phase is obtained.^ The resultant useful flux, 0, which induces 
this e.m.f. leads -B by 90 degree's in time; (P is in phase with the 
net or resultant m.m.f. Mn which produces it. The m.m.f. Mn is a 
sum of the field m.m.f. Mf and of the armature reaction Ma 

* On account of skin effect and eddy currents in the armature conductors, 
the effective resistance r to alternating currents is considerably higher than that 
calculated or measured with direct current. The actual amoimt of increase 
depends upon the character of the winding, the size of the conductors, the 
shape of the slots, the frequency, etc., so that no definite rule can be given. 
Fortunately, the ohmic drop constitutes but a small percentage of the voltage 
of a machine, so that a considerable error committed in estimating the value 
of the ir drop affects the voltage relations but very little. See A. B. Field, 
"Eddy Currents in Large Slot-wound Conductors," Trans. Amer, Inst, 
Elect, Engrs,, Vol. 24 (1905), p. 761. 



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Chap. VIII] REACTION IN SYNCHRONOUS MACHINES 



145 



expressed by eq. (64) . The triangle OFG represents the relations in 
space, while the figure OABD is a time diagram. Therefore, the 
two figures are independent of one another; but it is convenient 
to combine them into one, by using the common vectors and i. 




Fig. 37. — ^The perfonnance diagram of a synchronous generator ^ with 
non-salient poles. 

With respect to the triangle OFG, the vector i represents the posi- 
tion of the crest of the armature m.m.f . relatively to the crest OG 
of the field m.m.f., the angle between the two being 90 + </;, as is 
explained above. Thus, the vector Ma is in phase with i. 



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146 THE MAGNETIC CIRCUIT [Art. 47 

When i and e are given, the vector E is easily found if the 
resistance and the reactance of the armature winding are known. 
The required net excitation, Mn, is then taken from the no-load 
saturation curve of the machine, and Ma is figured out from eq. 
(64). Then the required field ampere-turns, M/, are found from 
the diagram, either graphically or analytically. 

The diagram shown in Fig. 37 is known as the Potier diagram. 
Strictly speaking, it is correct only for machines with non-salient 
poles, but as an approximate semi-empirical method it is some- 
times used for machines with projecting poles, in place of the more 
correct diagram shown in Fig. 40. Fig. 37 represents the condi- 
tions in the case of a generator with lagging currents. When the 
current is leading the vector i is drawn to the left of the vector e, 
with the corresponding changes in the other vectors. 

A similar diagram for a synchronous motor which draws a 
leading current from the line is shown in Fig. 38. The vector e' 
represents the line voltage, and e is the equal and opposite voltage 
which is the terminal voltage of the machine considered as a gen- 
erator. The rest of the diagram is the same as in Fig. 37. A lead- 
ing current with respect to the line voltage e' is a lagging current 
with respect to the generator terminal voltage e, so that the field 
is weakened by the armature reaction in both cases (M^<My.in 
both figures). The energy component ti of the current is reversed 
in the motor, therefore the field is shifted in the opposite direc- 
tion; Mn leads Mf in the motor diagram and lags behind it in the 
generator diagram. The case of a synchronous motor with a 
lagging current can be easily analyzed by analogy with the above- 
described cases. 

In practice, it is usually preferred to represent the relations 
shown in Figs. 37 and 38 analytically, rather than to actually con- 
struct a diagram. The following relations hold for both the gen- 
erator and the motor. Projecting all the sides of the polygon 
OABD on the direction e and on the direction perpendicular to 
and leading e by 90 degrees, we have 

E co8<f)g=e+irco&<f>+ixQin<f>, . . . (71) 
E siD.<l>g= ix cos <p —ir sin ^, .... (72) 

where ^« is the angle between the vectors e and Ey coimted positive 
when E leads e, as in Fig. 37. The subscript z suggests that the 



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Chap. VIII] REACTION IN SYNCHRONOUS MACHINES 



147 



angle (l)g is due to the impedance of the armature. The expressions 
i cos <l> and t sin ^ represent the energy component and the react- 
ive component of the current respectively; they are designated in 




Fig. 38. — ^The performance diagram of a synchronous motoTf with 
non-salient poles. 

Figs. 37 and 38 by ii and 12. Denoting the right-hand sides of the 
eqs. (71) and (72) by ei and 62 for the sake of brevity, we have: 



ei = e+iir+i2x; 
e2=i\x—i2T. 



(73) 
(74) 



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148 THE MAGNETIC CIRCUIT [Art. 47 

Squaring eqs. (71) and (72) and adding them together gives 

E^'^e^-Ve'^y (75) 

Dividing eq. (72) by (71) resiilts in 

tan <f)g= 62/61 (76) 

Consequently, the angle between E and i becomes known; namely, 

<l>' = <f>+<l>z, (76a) 

where ^' is called the internal phase angle. Knowing E, the cor- 
responding excitation Mn is taken from the no-load saturation 
curve of the machine; from the triangle OFG we have then: 

M/J=M„2 + Ma2+2M„Masin^', ... (77) 

where ^' is known from eq. (76a). In numerical applications it is 
convenient to express all the Af 's in kiloampere-tums. 

The diagram shown in Fig. 38 and the equations developed 
above can be used for determining not only the phase characteris- 
tics of a synchronous motor, but its overioad capacity at a given 
field current as well. This latter problem is of extreme importance 
in the design of synchronous motors. The input into the machine, 
per phase, is — ei cos <J); the part ir of the line voltage is lost in the 
armature, the part ix corresponds to the magnetic energy which is 
periodically stored in the machine and returned to the line, without 
performing any work. The remainder, E, corresponds to the use- 
ful work done by the machine, plus the iron loss and friction. If 
the armature possessed no resistance and no leakage reactance the 
terminal voltage would be equal to JE/ in magnitude and in phase 
position. Thus, the expression —Ei cos ^', corrected for the core 
loss in the armature iron, represents the input into the revolving 
structure, per phase. The overload capacity of the machine is 
determined by the possible maximum of this expression. 

The problem is complicated by the fact that the relation 
between E and Mn is expressed by the no-load saturation curve, 
which is difficult to represent by an equation. The problem is 

^ In numerical applications it is more convenient to use the approximate 
formula 

E'-e.+W/ei (75a) 

obtained by the binomial expansion of expression (75); since all other terms 
can be neglected when e, is small as compared to e^. 



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Chap. VIII] REACTION IN SYNCHRONOUS MACHINES 149 

therefore solved by trials, assuming a certain reasonable value of 
E, and calculating the expression — Kcos^', until a value of E 
is found, for which this expression, corrected for the core loss, fric- 
tion, and windage, becomes a maximum. The problem of finding 
i and ^' for an assumed E isa, definite one, because the four equa- 
tions (71), (72), (76a) and (77) contain only four unknown quanti- 
ties, i, ^, <l>zj and <J)\ Instead of solving the problem by trials, an 
analytical relation can be assumed between E and Af «, on the use- 
ful part of the no-load saturation curve, for instance a straight line 
(not passing through the origin), a parabola, etc. The problem is 
then solved by equating the first derivative of the product — 
Ei cos 0' to zero, having previously expressed E, t, and cos <{>' 
through some one independent variable. Both methods have 
been worked out for a synchronous motor with salient poles.^ The 
relations are simplified for a machine with non-salient poles. 

The foregoing theory of the armature reaction does not apply 
directly to single-phase machines. The pulsating armature reac- 
tion in such a machine can be resolved into two revolving reactions, 
as in Art. 42. The reaction which revolves in the same direction 
with the main field is taken into account as in a polyphase machine. 
The inverse reaction is partly wiped out by the eddy currents pro- 
"duced in the metal parts of the revolving structure ; it is therefore 
difficult to express the effect of this reaction theoretically. The 
treatment in this book is limited to polyphase machines, which are 
used in practice almost exclusively .^ 

Prob. 7. In the 100 kva., 440-volt, 6-pole, two-phase alternator, 
given in Problem 6, Art. 43, the amplitude of the first harmonic of the 
armature reaction was 4800(7« ampere-turns. What is the per cent 
voltage regulation of the machine at a power-factor of 80 per cent 
lagging, if Cs = l, that is if the armature has one conductor per slot? 
The armature reactance is 0.038 ohm, and the armature resistance is 
0.008 ohm, both per phase. The no-load saturation curve of the 
machine is as follows : 

e=400 440 490 525 550 volts. 
Mn =6.7 8.0 10.0 12.0 14.0 kiloamp.-tums. 

Ans. 22 per cent. 

* See the author's " Essays on Synchronous Machinery," Oeneral Electric 
Review y 1911, July and September. 

* In regard to the armature reaction in single-phase machines, see E. 
Arnold, Die Wechsdstromtechnik, Vol. 4 (1904), pp. 32-39; Pichelmayer, 
Dynamohau (1908), pp. 251-259; Max Wengner, Theoretische und Experir 
mentelle XJntersuchungen an der Synchronen Einphasen-Maschine (Oldenbourg, 
1911.) 



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150 THE MAGNETIC CIRCUIT [Art. 48 

Prob. 8. The machine specified above is to be used as a synchronous 
motor. Determine graphically the required field excitation when the 
useful output on the shaft is to be 700 kw., and in addition the machine 
must draw from the line 600 leading reactive kva. The eflficiency of the 
machine at the above-mentioned load is estimated to be about 91 per 
cent. Ans. 12.5 kiloampere-tums. 

Prob. 9. Draw to the same scale as the diagram shown in Fig. 37, 
another similar diagram, for the same value of the current and of the 
phase angle ^, except that the current is to be leading. Assume a 
reasonable shape of the saturation curve in determining the new value of 
Mn. Show that a much smaller exciting current is required with the 
same kva. output, than in the case of a lagging current.. 

Prob. 10. Solve problem 9 for the motor diagram shown in Fig. 38, 
assuming the current to be laggiug with respect to the line voltage. 

Prob. 11. For a given alternator, show how to determine the voltage 
e (Fig. 37), analytically or graphically, when Af/, i, and <f> are given; 
explain when such a c^se arises in practice. 

Prob. 12. For a given synchronous motor, show how to determine 
the reactive component ij of the current (Fig. 38), analytically or graphic- 
ally, when Mf, e and t'l are given ; explain when such a case arises in 
practice. 

Prob. 13. Work out the details of the above-mentioned method 
for the determination of the overload capacity of a synchronous motor 
by trials. Hint: Introduce the components of e and i, in phase and in 
quadrature with E; rewrite eqs. (71) and (72) by projecting the figure 
OABD on the direction of ^ and on that perpendicular to E. Use no 
angles in the formulae, and neglect the small terms containing r, where 
they lead to complicated equations of higher degrees. 

48. The Direct andTransverse Armature Reaction in a Synchro- 
nous Machine with Salient Poles. In a machine with non-salient 
poles the armature reaction shifts the field flux but hardly distorts 
its shape. In a machine with projecting poles the flux, generally 
speaking, is both altered in value and crowded toward one pole- 
tip (Fig. 36). It is convenient, therefore, to resolve the traveling 
wave of the armature m.m.f . into two waves, one whose crests coin- 
cide with the center lines of the poles, the other displaced by 90 
electrical degrees with respect to it. The first component of the 
armature m.m.f. produces only a " direct " effect upon the field 
flux, that is, it either strengthens or weakens the flux, without dis- 
torting it. The second component produces a " transverse " 
action only, viz., it shifts the flux toward one or the other pole-tip, 
without altering its value (that is, neglecting the saturation). 

We have seen before that an armature current, which reaches 
its maximum when the conductor is opposite the center of the 



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Chap. Vni] REACTION IN SYNCHRONOUS MACHINES 151 

pole, distorts the flux; while a current in quadrature with the 
former exerts a direct reaction only. It is natural, therefore, to 
resolve the actual current in each phase into two components, in 
time quadrature with each other, and in such a way that each 
component reaches its maximum in one of the above-mentioned 
principal positions of the conductor with respect to the field-poles. 
Let the current in each phase be i, and let it reach its maximum at 
an angle (p after the induced no-load voltage is a maximum (Fig. 
40) . Then, the two components of the current are 

id=i&m (p 
and 

i<=icos (p. 

The component i^ produces a direct armature reaction only, 
and the component it a transverse reaction only.^ 

For practical calculations, and in order to get a concrete picture 
of the armature reaction, it is convenient to represent the armature 
reaction as shown in Fig. 39. Namely, the direct reaction, due to 
the components i^ of the armature currents, is replaced by an equiv- 
alent number of concentrated ampere-turns M^ on the pole. The 
value of M^ is selected so that its action in reducing or strength- 
ening the flux is equal to the true action of the armature currents. 
The transverse reaction, due to the component it of the armature 
currents, is replaced by a certain number of ampere-turns, Mt, on 
the fictitious poles, (S), (iV), shown by dotted lines between the real 
poles. For simplicity, and for other reasons given in Art. 51, the 
fictitious poles are assumed to be of a shape identical with that of 
the real poles. The number of exciting ampere-turns Mt is so 
chosen, that the effect of the fictitious poles is approximately the 
same as that of the distorting ampere-turns on the armature. 

The flux of the fictitious poles strengthens the flux of the real 
poles on one side and weakens it by the same amount on the other 
side, so that the fictitious poles actually distort the main flux 
without altering its value. Strictly speaking, the complete action 
of the distorting ampere-turns on the armature cannot be imitated 

*The resolution of the armature reaction in a synchronous machine 
into a direct and a transverse reaction was first done by A. Blondel. See 
VIndustrie EUctrique, 1899, p. 481 ; also his book Moteurs Synchrones (1900), 
and two papers of his in the Trans. Intern, Electr. CongresSy St. Louis, 1904, 
Vol. 1, pp. 620 and 635. 



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152 



THE MAGNETIC CIRCUIT 



[Art. 48 



by fictitious poles of the same shape as the main poles, because 
harmonics of appreciable magnitude are thereby neglected. How- 
ever, actual experience shows that the performance of a machine, 
calculated in this way, can be made to check very well with the 
observed performance, by properly selecting the coefficients of the 
direct and the transverse reaction. In a generator, the flux is 
crowded against the direction of rotation of the poles (Fig. 36) ; 
consequently, the fictitiotis poles lag behind the real poles, as 
shown in Fig. 39. In a synchronous motor they lead the real 
poles by 90 electrical degrees. 

If the ratio of the pole arc to pole-pitch were equal to imity, 
as with non-salient poles, the whole wave of the demagnetizing 



Actual distribnfion 
of tFuwvene fflnx 



Fhix ^Istribafion 
doe to fictttioaa pole 




Fig. 39. — ^The direct and transverse armature reactions in a synchronous 
machine, represented by fictitious poles and field windings. 

m.m.f. of the armature would be acting upon the pole, and the 
equivalent concentrated m.m.f. M^ on the pole would have to be 
equal to the average value of the actual distributed armature 
m.m.f. We would have then 

ilfd=(2/7r)Msin^, (78) 

where the maximum armature m.m.f. is determined by eq. (64), 
Art. 43, and 2/;:= 0.637 is the ratio of the average to the maximum 
ordinate of a sine wave. In reality, only a part of the armature 
m.m.f ., the one near its amplitude, acts upon the poles, the action 
of lower parts of the wave being practically zero because of the 
gaps between the poles. Therefore, the ratio between the maxi- 
mum m.m.f. M sin ^ and the average equivalent m.m.f. Md is 



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Chap. Villi REACTION IN SYNCHRONOUS MACHINES 163 

larger than 0.637. For the ordinary shapes of projecting poles, 
experiment and calculation (see Art. 50 below) show that this 
ratio varies between 0.81 and 0.85. Using an average of these 
limits instead of 2/;r in eq. (78) and substituting for M its expres- 
sion from eq. (64) we obtain the following practical formula for 
estimating the armature demagnetizing ampere-turns per pole in a 
synchronous machine with projecting poles: 

Mrf=0.75fc6mmsin^ (79) 

In this formula i ^in is the component i^ of the armature current, 
per phase. In actual machines the numerical coefficient in this 
formula varies between 0.73 and 0.77, depending on the shape of 
the poles and the ratio of pole-arc to pole-pitch. 

By a similar reasoning, if the ratio of pole-arc to pole-pitch were 
equal to unity, the equivalent number of exciting ampere-turns on 
the fictitious poles would be 

Mt=(2/7r)Mcos0 (80) 

Since the ratio of pole-arc to pole-pitch on the fictitious poles 
is less than irnity, the numerical coefficient should be larger than 
2/;r. But, on the other hand, the permeance of the air-gap imder 
the fictitious poles is much higher than the actual permeance of 
the machine in the gaps between the poles, so that a much smaller 
number of ampere-turns Mt is sufficient to produce the same 
distorting flux. The combined effect of these two factors is to 
reduce the coefficient in formula (80) to a value considerably 
below 2/n. For the usual shapes of projecting poles, experiment 
and calculation (See Art. 51 below) show that this ratio varies 
between 0.30 and 0.36. Using an average of these limits instead 
of 2/n in eq. (80), and substituting for M its expression from eq. 
(64), we obtain the following practical formula for estimating 
the distorting ampere-turns per pole, in a synchronous machine 
with projecting poles: 

Mt=^0.30ki,mnicoa(lf (81) 

In this formula i cos ^ is the component it of the armature cur- 
rent, per phase. In some actually built machines the coefficient 
in this formula comes out lower than 0.30, but in preliminary cal- 



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154 



THE MAGNETIC CIRCUIT 



[Art. 49 



culations it is advisable to use at least 0.30. When a synchronous 
motor is working near the limit of its overload capacity, the influ- 
ence of the distorting ampere-turns is particularly important, and 
in estimating the overload capacity of a synchronous motor it is 
better to be on the safe side and to take the value of the numerical 
coeflScient in eq. (81) somewhat higher than 0.30. The value of 
this coeflScient varies within wider limits than that of the corre- 
sponding coefficient 
in formula (79); 
but, fortunately, it 
affects the perform- 
ance to a lesser de- 
gree (see Art. 51). 

49. TheBlondel 
Performance Dia- 
gram of a Syn- 
chronous Machine 
with Salient Poles. 
Having replaced the 
actual armature 
reaction by two 
m.m.fs. M^ and Mt 
(Fig. 39) the elec- 
tromagnetic rela- 
tions in the machine 
become those indi- 
cated in Figs. 40 and 
41. Fig. 40 refers 
to a generator and 
is analogous to Fig. 
37; Fig. 41 refers 
to a motor and is 
analogous to Fig. 
38. The polygon OABD, which represents the relation between 
the terminal and the induced voltages, is the same as before, but 
the induced voltage E is now considered as a resultant of the 
voltages En and Et induced by the real and the fictitious poles 
respectively.^ In the generator the fictitious poles lag behind 

^ The subscript n stands for net, to agree with the m.m.f . Mn used later 
on; the subscript t stands for transverse. 




Fig. 40. — The performance diagram of a synchro- 
nous generator, with salient poles. 



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Chap. VIII] REACTION IN SYNCHRONOUS MACHINES 



155 



the real ones, in a motor they lead the real poles. Hence, in the 
generator diagram, Et lags 90 degrees behind En, while in the 
motor diagram it leads En by 90 degrees. 

In the case of a generator the problem usually is to find the 
field excitation Af/ neces- 
sary for maintaining a ^ 
required terminal voltage ^ 
6, with a given current i 
and at a given power- 
factor cos <f). First, the 
figure OABD is con- 
structed, or else the 
values of E and ^' are 
determined from eqs. 
(75), (76), and (76a). 
In order to find the 
ampere-turns required on 
the main poles it is neces- 
sary to determine the 
voltage En induced by 
them. For this purpose 
the angle /? must first be 
known, for 

En= E cos^. . (82) 

As an intermediate step, 
it is necessary to express 
Et through the ampere- 
turns Mt, which are the 
cause of Et. The m.m.f . 
Mt is small as compared 
to the total number of 
ampere-turns on the real 
poles; hence, the lower 
straight part of the no- 
load saturation curve of 
the machine can be used 
to express the relation 
between Mt and Et. Let v be the voltage corresponding to 
one ampere-turn on the lower part of the no-load saturation 




Fig. 41. — ^The performance diagram of a syn- 
chronous motor y with salient poles. 



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166 THE MAGNETIC CIRCUIT [Art. 49 

curve; then Et =MtV. Substituting the value of Mt from eq. (81), 
we have 

St=-E/cos((^'+/?), (83) 

where 

E/ =-0.30hmniv (84) 

E/ is a known quantity introduced for the sake of brevity. The 
angle ^ in formula (83) is expressed through ^' and /?, because, 
from Fig. 40, 

0=(^'+/? (85) 

Another relation between Et and /? is obtained from the triangle 
ODGj from which 

Et=E sin p (86) 

A comparison of eqs. (83) and (86) gives that 

E/co8(<J>'+p) = Eamp. 

Expanding and dividing throughout by cos p we find the relation 



cos^' 
(B/ETT+sm ^' 



^^^P-7W7irK~:-rTr' ^^7) 



from which the angle /? can be determined, and then E^ calculated 
byeq. (82). 

The next step is to take from the no-load saturation curve the 
value Mn of the net excitation necessary on the main poles in order 
to induce the voltage E^. The real excitation Mf must be larger, 
because part of it is neutralized by the direct armature raction Af d. 
We thus have 

Mf=Mn+Ma, (88) 

where Md is calculated from eq. (79), the angle <p being known 
from eq. (85). When the load is thrown ofif, the only excitation 
left is Mf'j let it correspond to a voltage e© on the no-load satura- 
tion curve. From e© and e the per cent voltage regulation of the 
machine is determined from its definition as the ratio (e©— e)/e. 

The same general method and the same equations apply in the 
case of Fig. 41, when one is required to determine a point on one of 
the phase characteristics of a synchronous motor. The beginner 
mtist be careful with the sign minus in the case of the motor. 



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Chap. VIII] REACTION IN SYNCHRONOUS MACHINES 157 

Since ^' > 90 degrees, the angle ^ and the voltage Et are negative. 
The angle <l>g also is usually negative. The cases of a leading 
current in the generator and of a lagging current in the motor are 
obtained by assigning the proper value and sign to the angle <j>. 
For the application of the Blondel diagram to the determination 
of the overload capacity of a synchronous motor see the reference 
given near the end of Art. 47. 

A synchronous motor is sometimes operated at no load, and at 
such a value of the field current that the machine draws reactive 
leading kilovolt-amperes from the line, thus improving the 
power-factor of the system. In such a case the machine is called a 
synchronous condenser , or better, a phase adjuster. The diagram in 
Fig. 41 is greatly simplified in this case because the energy com- 
ponent of the current can be neglected, as well as the drop ir, 
and the e.m.f. Et. We then have i=t2 = H and E^=E=e+ix, 
The direct armature reaction is determined from eq. (79) in which 
^=90. When the motor is imderexcited and draws a lagging 
current from the line, i is to be considered negative, or ^=270 
degrees. The same simplified diagram applies to a polyphase 
rotary converter, operated from the alternating-current side, at no 
load. 

Prob. 14. It is required to calculate the field current and per cent 
voltage regulation of a 12-pole, 150 kva., 2300- volt, 60-cycle, Y-connected 
alternator, at a power factor of 85 per cent lagging. The machine has 
two slots per pole per phase, and is provided with a full-pitch winding, 
the number of turns per pole per phase being 18. The armature resist- 
ance per phase of Y is 0.67 ohm, the reactance is 3.5 ohm. The number 
of field turns per pole is 200. The no-load saturation curve is plotted 
for the line voltage (not the phase voltage), and at first is a straight line 
such that at 1800 volts the field current is 17.4 amp. The working part 
of the no-load saturation curve is as follows : 

Kilovolts 2.2 2.4 2.5 2.6 2.7 2.78 

Field current, amp 22 25 27 30 34 40 

Ans. 31 amp. ; 14.3 per cent. 

Pnob. 16. Show that in the foregoing machine the short-circuit 
current is equal to about two and a half times the rated current, at the 
field excitation which gives the rated voltage at no-load. Hint: The 
short-circuit curve is a straight line so that one can first calculate the 
field current for any assumed value of the armature current and c =0. 

Prob. 16. From the results of the calculations of the preceding 
problem show that the cross-magnetizing effect and the ohmic drop are 
negligible under short-circuit, in the machine under consideration. 



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158 THE MAGNETIC CIRCUIT [Art. 60 

Assuming that r is usually small as compared to x, describe a simple 
method for calculating the short-circuit curve, using only the reactance 
of the machine and the demagnetizing ampere-tums of the armature. 
In practice, the influence of the neglected factors is accounted for in short- 
circuit calculations by taking sin (p in formula (79) as equal to between 
0.95 and 0.98 instead of unity. 

Prob. 17. Plot the no-load phase characteristic of the machine 
specified in problem 14, when it is used as a motor. The iron loss and 
friction amount to 8.5 kw. 

Ans. Field amperes 14.9 23.4 32.6 

Armature amperes 30 2.13 30 

Prob. 18. The machine specified in problem 14 is to be used as a 
motor, at a constant input of 150 kw. Plot its phase characteristics, 
i.e., the curves of the armature current and of power-factor against the 
field current as abscissae. 

Ans. Field amperes 32.6 24.3 16.65 

Armature amperes.. 47.00 37.65 47.00 
Power-factor 0.80 1 .00 0.80 

Prob. 19. Write complete instructions for the predetermination of 
the regulation of alternators and of the phase characteristics of synchro- 
nous motors, by BlondePs method. The instructions must give only 
the successive steps in the calculations, without any theory or explana- 
tions. Write directions and formulae on the left-hand side of the sheet, 
and a numerical illustration on the right-hand side opposite it. 

Prob. 20. Calculate the overload capacities of the foregoing motor 
at field currents of 25 amp. and 35 amp., by the two methods described 
in the articles refered to near the end of Art. 47. 

Prob. 21. Show that for a machine with non-salient poles Blondel's 
and Potier's diagrams are identical. 

60. The Calculation of the Value of the Coefficient of Direct 
Reaction in Eq. (79) .^ The average value 0.83 of the ratio of the 
effective armature m.m.f . over a pole-face to the maximum m.m.f. 
at the center of the pole is given in Art. 48 without proof. The 
following computations show the reasonable theoretical limits of 
this ratio. If the armature m.m.f. (direct reaction) at the center 
of the N pole (Fig. 39) is Af, its value at some other point along 
the air-gap is M cos x, where x is measured in electrical radians. 
Let the permeance of the active layer of the machine per electrical 
radian be (P at the center of the pole, and let this permeance vary 
along the periphery of the armature according to a law /(x), so 
that at a point determined by the abscissa x the permeance per 

' This and the next article can be omitted, if desired, without impairing 
the continuity of treatment. 



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Chap. VIU] REACTION IN SYNCHRONOUS MACHINES 159 

electrical radian is (Pf{x). The function /(a:) must be periodic and 
such that /(O) = 1, and fUn) =0, /(;:) = 1, etc., because the perme- 
ance reaches its maximum value \mder the centers of the poles 
and is practically nil midway between the poles. 

The direct armature m.m.f., acting alone, without any excita- 
tion on the poles, would produce in each half of a pole a flux 



M cos X (Pf{x)dx, 



The magnetomotive force Md placed on the real poles, acting 
alone, must produce the same total flux, so that 

0=Maf^^^yfix)dx. 

Equating the two preceding expressions we get 

MC^\o8xf{x)dx^MdCJy{x)dx. . . . (89) 

The ratio of Md to M can be calculated from this equation, by 
assuming a proper law f{x) according to which the permeance of 
the active layer varies with x, in poles of the usual shapes. Hav- 
ing a drawing of the armature and of a pole, the magnetic field can 
be mapped out by the judgment of the eye, assisted if necessary 
by Lehmann's method (Art. 41 above). A curve can then be 
plotted, giving the relative permeances per imit peripheral length, 
against x as abscissae. Thus, the fimction/(x) is given graphically, 
and the two integrals which enter into eq. (89) can be determined 
graphically or be calculated by Simpson's Rule. Or else, 
f(x) can be expanded into a Fourier series and the integration 
performed analytically. Such calculations performed on poles of 
the usual proportions give values of Md/M of between 0.81 and 
0.85. 

It is also possible to assume for f(x) a few simple analytical 
expressions, and integrate eq. (89) directly. Take for instance 
fix) =cos2 X, By plotting this fimctipn against x as abscissae the 
reader will see that the function becomes zero midway between 
the poles, is equal to unity opposite the centers of the poles, and 
has a reasonable general shape at intermediate points. Substi- 
tuting cos^ X for f(x) into eq. (89) and integrating, gives f M = 
i7:Md, from which Md/M =0.85. 



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160 . THE MAGNETIC CIRCUIT [Art. 51 

Another extreme assumption is that of poles without chamfer, 
with a constant air-gap. Neglecting the fringe at the pole-tips, 
/(x) = 1 from X =0 to X =5, and /(x) =0 from x ==5 to x =^7:. Inte- 
grating eq. (89) between the limits and we obtain 

Mrf/M = (sin»)/» (90) 

The poles usually cover between 60 and 70 per cent of the periph- 
eiy. For =0.6(i;r) the preceding equation gives Md/M =0.86, 
and for 0=O.7(i;r), Mrf/M=0.81. 

Prob. 22. Let the permeance of the active layer decrease from the 
center of the poles according to the straight-line law, so that 

/(x)=l-(2A)x. 

What is the ratio of MdlMf Ans. 0.811. 

Prob. 23. The permeance of the active layer decreases according 
to a parabolic law, that is, as the square of the distance from the center 
of the poles. What is the ratio of Md/Mf Ans. 0.774. 

Prob. 24. The law /(x) =cos* x assumed in the text above presup- 
poses that the permeance varies according to a sine law of double 
frequency with a constant term, because cos* x =i + J cos 2x. In reality, 
the permeance varies more slowly under the poles and more rapidly 
between the poles than this law presupposes (Fig. 39). A correction can 
be brought in by adding another harmonic of twice the frequency to the 
foregoing expression, thus making it unsymmetrical, and of the form 
/(x) =a + 6 cos 2x4- c cos 4x. Show that /(x) =2 cos* x— cos* x contains 
the largest relative amount of the fourth harmonic, consistent with the 
physical conditions of the problem, and compare graphically this curve 
with /(x)= cos* X. 

Prob. 26. What is the value of Md/M for the form of /(x) given in 
the preceding problem? Ans. 0.815. 

Prob. 26. Plot the curve /(x) for a given machine, estimating the 
permeances by Lehmann's method, and determine the value of the coef- 
ficient in formula (79). 

61. The Calculation of the Value of the Coefficient of Transverse 
Reaction in Eq. (81). The average value 0.33 of the ratio of the 
maximum distorting armature m.m.f . to the equivalent number of 
ampere-turns, M^, on the fictitious poles is given in Art. 48 without 
proof. The following computations show the reasonable theoret- 
ical limits of this ratio. The problem is more complicated than 
that of finding the ratio of Md/M, because there the field ampere- 
turns, the actual demagnetizing armature-m.m.f., and the equiva- 
lent ampere-turns Md are all acting on the same permeance of the 



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Chap. VIII] REACTION IN SYNCHRONOUS MACHINES 161 

active layer, and the wave form of the flux is very little affected by 
the direct armature reaction. In the case of the transverse reac- 
tion, however, the wave form of the flux produced by the actual 
cross-magnetizing ampere-turns of the armature is entirely differ- 
ent from that produced by the coil Mt acting on the fictitious pole 
(Fig. 39). Namely, the actual curve of the transverse flux has a 
large " saddle " in the middle, due to the large reluctance of the 
space between the real poles. The flux distribution produced by 
the fictitious poles is practically the same as that imder the main 
poles, the two sets of poles being of the same shape. 

The addition of the vectors Et and E^ in Figs. 40 and 41is legit- 
imate only when Et is induced by a flux of the same density dis- 
tribution as E^, and this is the reason for representing the trans- 
verse reaction as due to fictitious poles of the same shape as the real 
poles. Therefore, for the purposes of computation, the flux dis- 
tribution, produced by the actual distorting ampere-turns on the 
armature, is resolved into a distribution of the same form as that 
produced by the main poles and into higher harmonics. The m.m.f . 
Mt is calculated so as to produce the first distribution only. This 
fundamental curve is not sinusoidal, but will have a shape depend- 
ing on the shape of the pole shoes. The effect of the sinusoidal 
higher harmonics on the value of Et is disregarded, or it can be 
taken into accoimt by correcting the value of the coefficient in 
formula (81) from the results of tests. 

The first harmonic of the armature distortion m.m.f. is M sin x, 
because this m.m.f. reaches its maximum between the real poles; 
X is measured as before from the centers of the real poles. The 
permeance of the active layer, with reference to the real poles, can 
be represented as before by (Pfix). The flux density produced by 
the transverse reaction of the armature at a point defined by the 
abscissa x is therefore proportional to ilfsinx (Pf(x). The per- 
meance of the active layer with reference to the fictitious poles is 
(Pf(x+in). The flux density imder the fictitious poles follows 
therefore the law Mt(Pf(x+in), As is explained before, the two 
distributions of the flux density differ widely from one another, 
and the real distribution is resolved into the fictitious distribution, 
and higher sinusoidal harmonics; 'the prominent third harmonic is 
clearly seen in Fig. 39. Thus, we have, omitting (P, 

M sin X fix) = Mtfix + J;r) + ^1 3 sin 3a: + 4 5 sin 5x + etc. (91) 

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162 THE MAGNETIC CIRCUIT [Art. 51 

In order to determine Mt, the usual method is to mulitply both 
sides of this equation by sin x and integrate between and n^ 
because then all upper harmonics give terms equal to zero. In 
this particular case the limits of integration can be narrowed down 
to and ^n, because the symmetry of the curve is such that the 
segment between and \n is similar to all the rest. Thus, we 
get 

mC\vd? xf{x)dx =M< r*"sin xf(x+in)dx. . (92) 

From this equation the ratio Mt/M can be calculated by the 
methods shown in Art. 50, i.e., by assuming reasonable forms of 
the fimction /(x). Taking again f(x) =cos2 x and integrating eq. 
(92) we get J^;rM=JM<, from which M</Af =0.295. Taking the 
other extreme case, viz., f(x) =1 from x=0 to x=0, and f(x) =0 
from a; =5 to a: =i;r, gives, after integration 

Mt/M = {id-iQm2e\/Qiiie (93) 

For (?=0.6(i;r), M</M =0.29; for 5=0.7 (i;r), ilf«/ilf=0.39.i It 
will be noted that the cross-magnetizing action of the armature 
increases considerably with the increasing ratio of pole-arc to pole- 
pitch, while the direct reaction slowly diminishes with the increase 
of this ratio. In machines intended primarily for lighting pur- 
poses it is advisable to use a rather small ratio of pole-arc to pole- 
pitch, in order to reduce transverse reaction which affects the volt- 
age regulation at high values of power-factor in particular. 

Prob. 27. What is the value of Mt/M for the form of f(x) given in 
problem 24; namely, for/(x) = 2 cos^ a; -cos* x? Ans. 0.368. 

Prob. 28. Determine the numerical value of the coefficient in formula 
(81) for the machine used in problem 26. 

^ These values are higher than those derived by E. Arnold. . The fact 
that Arnold's values for the coefficient of transversal reaction are low has 
been pointed out by Sumec in Elektrotechnik und Maschinenbau^ 1906, p. 
67; also by J. A. Schouten, in his article " Ueber den Spannungsabfall mehr- 
phasiger synchroner Maschinen, "EUktrotechnische ZeUachrift^ Vol. 31 (1910), 
p. 877. 



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CHAPTER IX 

ARMATURE REACTION IN DIRECT-CURRENT 
MACHINES 

62. The Direct and Transverse Armature Reactions. Let 

Fig. 42 represent the developed cross-section of a part of a direct- 
current machine, either a generator or a motor. For the sake of 
simplicity the brushes are shown making contact directly with the 
armature (conductors, omitting the commutator. Electrically 
this is equvialent to the actual conditions, because the commutator 
segments are soldered to the end-connections of the same conduc- 
tors. The brushes are shifted by a distance d from the geometrical 
neutral, to insure a satisfactory commutation; d being expressed 
in centimeters, measured along the armature periphery, the same 
as the pole-pitch t. 

The actual armature conductors and currents are replaced, for 
each pole-pitch, by a current sheet, or belt, of the same strength. 
Let, for instance, the pole-pitch be 40 cm., and let the machine 
have 120 armature conductors per pole. If the current per con- 
ductor is 100 amp., the total number of ampere-conductors per 
pole is 12,000; the total current of the equivalent belt, which con- 
sists of one wide conductor, must be 12,000 amp., or 300 amp. per 
linear cm. of the pole-pitch. The latter value, or the number of 
armature ampere-conductors per centimeter of periphery, is some- 
times called the specific electric loading of the machine. The mag- 
netic action of the equvialent current sheet on the magnetic flux of 
the machine is practically the same as that of the actual armature 
conductors, because in a direct-current machine the slots are com- 
paratively numerous and small. The current in the cross-hatched 
belts is supposed to flow from the reader into the paper, and the cur- 
rent in the belts marked with dots — ^toward the reader. With the 
directions of {he flux and of the current shown in the figure, the 
directions of rotation of the machine when working as a generator 
and as a motor are those shown by the arrow-heads (see Art. 24). 

163 



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164 



THE MAGNETIC CIRCUIT 



[Aet. 52 



The polarity of the brushes cannot be indicated without knowing 
the actual connections in the winding. It is preferable, therefore, 
for our purposes to designate the brushes as E and W (east and 
west), according to their position with respect to the poles of the 
machine, the observer looking from the commutator side. The 
whole interpolar regions to the right of the north poles can be called 
the eastern regions, those to the left the western regions; the same 
notation can be also applied to the commutating poles. 

The armature currents exert a two-fold action upon the main 
field of the machine : they partly distort it, and partly weaken it. 
For the purposes of theory and calculation it is convenient to 
separate these two actions, the same as in the case of the synchro- 




FiQ. 42. — ^The direct and transverse armature reactions in a 
direct-current machine. 

nous machine in the preceding chapter. Let the sheets of current 
be divided into parts denoted by the letters D and T with sub- 
scripts corresponding to their location with reference to the poles 
and brushes. The belts denoted by D are comprised within the 
space dj one each side of the geometrical neutrals; those denoted 
by T are {^r—d) centimeters wide. 

The belts D exert a direct demagnetizing action upon the poles. 
Namely, the belts Dn Dn can be considered as two sides of a coil the 
axis of which is along the center line CnCn. The m.m.f. of this 
coil opposes that of the field coil on the north pole. In the same 
way, the m.m.f. of the coil DaDg opposes! the action of the field coil 
on the south pole. The foregoing is true no matter what the 
actual connections of the armature conductors are, provided that 
the winding-pitch is nearly 100 per cent. With a fractional-pitch 



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Chap. IX] ARMATURE REACTION IN D.C. MACHINES 165 

winding the currents within each D belt flow partly in the opposite 
directions and neutralize each other's action. 

Let the specific electric loading of the machine, as defined 
above, be (AC). Then, with a full-pitch winding, the demagnetiz- 
ing ampere-turns per pole are^ 

Mi = (4C)5 . (94) 

The belts TeTe constitute together a coil the center of which is 
along the axis OeOe\ the adjacent belts jT^r^form a coil with its 
axis along OJOJ. The m.m.f. distribution of these coils is indi- 
cated by the broken line ABC, which shows that the T belts pro- 
duce a transverse armature reaction. The line ABC is also the 
curve of the flux density distribution which would be produced by 
the transversal reaction alone, if the active layer of the machine 
were the same throughout (non-salient poles). On accoimt of a 
much higher reluctance of the paths in the interpolar regions the 
flux density there is much lower, and is shown by the dotted lines. 
The actual distribution of the field in a loaded machine is obtained 
considering from point to point the field and armature m.m.fs. 
acting upon the individual magnetic paths. 

The transverse reaction opposes the field m.m.f. imder one-half 
of each pole and assists it under the other half, so that the main 
field is distorted. In a generator the field is shifted in the direc- 
tion of rotation, in a motor it is crowded against the direction of 
rotation of the armature. This is the same as what happens in 
synchronous machines, when the armature is revolving and the 
poles stationary (see Fig. 36). 

The brushes must be shifted in the same direction in which the 
flux is shifted, because the magnetic neutral is displaced with 
respect to the geometric neutral. Usually, the brushes are shifted 
beyond the magnetic neutral, in order to obtain a proper flux den- 
sity for commutation. Namely, to assist the reversal of the cur- 
rent in the conductors which are short-circuited by the brushes, 
these conductors must be brought into the fringe of a field of such 
a direction as assists the commutation. In the case of a gener- 
ator this means the field imder the influence of which the conduc- 

^ The effect of the coils short-circuited under the brushes is not considered 
separately, for the sake of simplicity. For an analysis of the reaction of the 
short-circuited coils upon the field see E. Arnold, Die Gleichstrommaschine^ 
Vol. 1 (1906), Chap. 23. 



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166 THE MAGNETIC CIRCUIT IArt. 53 

tors come after the commutation. In a motor the armature cur- 
rent flows against the induced e.m.f.; it is therefore the field which 
cuts the conductors before the commutation that assists the rever- 
sal of the current. This explains the direction of the shift of the 
brushes in the two cases. The student should make this clear to 
himself by considering in detail the directions of the currents and 
of the induced voltages in a particular case. 

The maximum m.m.f. per pole produced by the distorting 
belts is equal to {AC) (^r— <?), but since this m.m.f. acts along the 
interpolar space of high reluctance its effect is not large (except in 
machines with commutating poles). Of much more importance 
is the action of the distorting belts under the main poles. At each 
pole-tip the armature m.m.f. is 

M2 = (ilC)4^, (95) 

where w is the width of the pole shoe. This m.m.f. decreases 
according to the straight line law to the center of each pole and 
is of opposite signs at the two tips of the same pole. 

Prob. 1. Determine the polarity of the brushes in Fig. 42 for a 
progressive and a retrogressive winding, in the case of a generator and 
of a motor. 

Prob. 2. Indicate the D and the T belts in a fractional-pitch winding 
(a) with the brushes in the geometric neutral, and (6) with the brushes 
shifted by d, 

Prob. 3. A 500 kw., 230 volt, 10 pole, direct-current machine has 
a full-pitch multiple winding placed in 16S^ slots. There are 8 con- 
ductors per slot, and two turns per commutator segment. What are 
the demagnetizing ampere-turns per pole when the brushes are shifted 
by 4 commutator segments? Ans. 3470. 

Prob. 4. What is the amplitude of the broken line ABC in the 
preceding machine? Ans. 10850 amp-turns. 

Prob. 6. For a given machine, draw the curves of the flux density 
distribution under a pole, at no-load and at full load, by considering 
the m.m.fs. acting upon the individual paths, and the reluctance of the 
paths. ^ 

63. The Calculation of the Field Ampere-turns in a Direct- 
current Machine under Load. The net ampere-turns per pole are 
determined from the no-load saturation curve of the machine for 

* For details and examples of such curves see Pichelmayer, Dynamobau 
(1908), p. 176; Parshalland Hobart, Electric Machine Design (1906), p. 159; 
Arnold, Die Gleichstrommachinej Vol. 1 (1906), p. 324. 



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Chap. IX] ARMATURE REACTION IN D.C. MACHINES 167 

the necessary induced voltage. In a generator the induced voltage 
is equal to the specified terminal voltage plus the internal ir drop 
in the machine. In a motor the induced voltage is less than the 
line voltage by the amount of the internal voltage drop. The 
actual ampere-turns to be provided on the field poles are larger 
than the net excitation by the amoimt necessary for the compensa- 
tion of the armature reaction. 

The direct reaction is compensated for by adding to each field 
coil the ampere-turns given by eq. (94). For instance, in 
prob. 3 above, 3470 ampere-turns per pole must be added to the 
required net excitation, in order to compensate for the effect of the 
direct armature reaction. The necessary shift of the brushes is 
only roughly estimated from one's experience with previously 
built machines, though it could be determined more accurately 
from the distribution of flux density in the pole-tip fringe. The 
poles usually cover not over 70 per cent of the armature periphery, 
so that the distance between the geometric neutral and the pole- 
tip is about 15 per cent of the pole pitch. In preliminary esti- 
mates, the brush stdf t may be taken to be about 10 per cent of the 
pole pitch; this brings the brushes not quite to the pole-tips 
though well within their fringe. In actual operation a smaller 
shift may be expected. In machines provided with commutating 
poles, and in motors intended for rotation in both directions, the 
brushes are usually in the geometric neutral, so that the demag- 
netizing action is zero. 

In a machine with a low saturation in the teeth and in the pole- 
tips, the cross-magnetizing m.m.f. of the armature does not affect 
the magnitude of the total flux per pole, because the flux is 
increased on one-half of the pole as much as it is reduced on the 
other half. It is shown in Art. 31 that the induced e.m.f. of a 
direct-current machine is independent of the flux distribution, 
provided that the total flux is the same, so that no extra field 
ampere-turns are necessary in such a machine to compensate for 
the distortion of the flux. 

However, the teeth and the pole-tips are usually saturated to 
a considerable extent, so that the flux is increased on one side of 
the pole less than it is reduced on the other side. Thus, the useful 
flux is not only distorted by the transverse armature reaction, but 
is also weakened. This latter effect has to be counterbalanced by 
additional ampere-turns on the field poles. In most cases these 



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168 



THE MAGNETIC CIRCUIT 



[Art. 53 



additional ampere-turns are estimated empirically, on the basis 
of one's previous experience, because the amount is not large, and 
a correct computation is rather tedious.^ 

The theoretical relation between the distorting ampere-turns 
and the field ampere-turns required for their compensation is shown 
in Fig. 43. Let OX represent the no-load saturation curve of the 
machine for its active layer only, that is, for the air-gap, the teeth, 
and the pole shoe, if the latter is sufficiently saturated. The 
ordinates represent the induced e.m.f . between the brushes, or, to 
another scale, the useful flux per pole; the abscissae give the corre- 
sponding values of the field ampere-turns per pole, disregarding 
the reluctance of the field poles, field yoke, and armature core. 



hhL 




Fig. 43. — A construction for determining the field m.m.f. needed for the 
compensation of the transverse reaction. 

In other words, the abscissae give the values of the difference of 
magnetic potential across the active layer of the machine, at no 
load. It is proper to consider here the m.m.fs. across the active 
layer only, because the distorting action of the armature extends 
only over this layer. No matter how irregular the flux distribu- 
tion in the air-gap and in the teeth may be, the flux density in the 
pole cores and in the yoke is practically \miform (compare Fig. 36). 

^ For Hobart's empirical curves for estimating the field excitation required 
for overcoming the armature distortion see the Standard Handbook, under 
* Generators, direct-current, ampere-tums, estimate." 



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Chap. IX] ARMATURE REACTION IN D.C. MACHINES 169 

For the sake of simplicity, we replace the actual pole shoe by 
an equivalent one, without chamfer, of rectangular shape, and 
with a negligible pole-tip fringing. The ordinates of the curve OX 
represent to another scale the average flux density on the surface 
of the pole shoe, because this density is proportional to the total 
useful flux, or to the induced voltage. Let ab be the flux density 
corresponding to the required induced e.m.f., and let Oa=M be 
the necessary net m.m.f ., without the transverse reaction. Let ac 
and dd represent the distorting ampere-turns, Af 2, at the pole-tips, 
as given by eq. (95) ; then Oc and Od are the resultant m.mis. at 
the pole-tips. The flux density at the pole-tips of the loaded 
machine is then equal to eg and dh respectively. 

Since the distorting ampere-turns vary directly as the distance 
from the center of the pole, and the area of strips of equal width 
is the same, the abscissae from a represent to scale either distances 
along the pole face, or areas on the pole face, measured from the 
center of the pole. Thus, the part gh of the curve OX represents 
also the distribution of the flux density imder the pole, in the 
loaded machine. Likewise, the line ef represents the distribution 
of the flux density imder the pole at no-load. 

Since the ordinates represent to scale the flux densities and the 
abscissae the areas of the different parts imder the poles, the area 
imder the flux density distribution curve also represents to scale 
the total flux per pole. The total flux at no load is represented by 
the area of the reactangle cefd, and to the same scale, the flux in the 
loaded machine is represented by the area cghd. If the saturation 
curve were a straight line, the two areas would be equal, so that 
the distortion would not modify the value of the total flux per pole. 
In realityi, the area geb is larger than the area bhf, and the differ- 
ence between the two represents the reduction in the flux, due to 
the transverse armature reaction. 

Let now the field excitation be increased by an imknown amount 
aa' to the value Oa^ =Af' , in order to compensate for the above 
explained decrease in the flux. All the points in Fig. 43 are shifted 
by the same amount, and are denoted by the same letters with the 
sign " prime." The new flux in the loaded machine is represented 
by the area dg'Kd'j the point a! being the center of the line c'& =cd. 
If the point a' has been properly selected we must have the con- 
dition that the 

area cefd =area c'g^Wd', (96) 



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170 THE MAGNETIC CIRCUIT [Art. 53 

that is, the flux corresponding to the excitation Oa at no-load is 
the same as the flux corresponding to the excitation Oa when the 
machine is loaded. The problem is then, knowing the point a and 
the distance cd =c'd' to find the point a' such that equation (96) 
is satisfied. The two areas have the common part c'g'hfdj and 
the parts cee'c' and dff'd' are equal to each other. Therefore, 
eq. (96) is satisfied if the 

area6ey=area6/'A' (97) 

The position of the point a' is found either by trials, or as the 
intersection of the curves rr' and ss'; the ordinates of the curve rr^ 
represent the area be'g' for various assumed positions of the point 
a', and the ordinates of the curve ««' represent the corresponding 
values of the area bfW. 

Thus, the total field ampere-turns required for a direct-current 
generator imder load are found as follows: (a) To the specified 
terminal voltage add the voltage drop in the armature, under the 
brushes, and in the windings (if any) which are in series with the 
armature, viz. ; the series field winding, the compensating winding, 
and the interpole winding. This will give the induced voltage E. 
(6) From the no-load saturation curve find the excitation corre- 
sponding to E, (c) Estimate the amoimt of the brush shift (if 
any) and calculate the corresponding demagnetizing ampere-turns 
according to eq. (94) . (d) Determine the ampere-turns aa' (Fig. 43) 
required for the compensation of the armature distortion, (e) 
Add the ampere-turns calculated under (6), (c) and id). In the 
case of a motor subtract the voltage drop in the machine from 
the terminal voltage, to find the induced e.m.f. E, but otherwise 
proceed as before. 

If the machine is shxmt-woimd or series-woimd, the field wind- 
ing is designed so as to provide the necessary maximum number of 
ampere-turns at a required margin in the field rheostat for the 
specified voltage or speed variations. When the machine is com- 
poimd woimd, the shimt winding alone must supply the required 
number of ampere-turns at no-load. The series ampere-turns is, 
then, the difference between the total m.m.f. required at full-load 
and that supplied by the shunt-field. When a generator is over- 
compoimded, the sh\mt excitation is larger at full load than it is at 
no load, and allowance must be made for this fact. 



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Chap. IX] ARMATURE REACTION IN D.C. MACHINES 171 

In the case of a variable-speed motor, the problem of predict- 
ing the exact speed at a given load can be solved by successive 
approximations only. First, this speed is determined neglecting 
the transverse armature reaction altogether, or assigning to it a 
reasonable value. Then, the construction shown in Fig. 43 is 
performed for a few speeds near the approximate value, and thus 
a more correct value of the speed is found by trials. Or else 
different values of speed are assumed first, and the corresponding 
values of the armature current are found for each speed. These 
armature currents must be such that, considering the total arma- 
ture reaction, the field m.m.f. is just suflBcient to produce the 
required coimter-e.m.f. in the armature. The details of the solu- 
tion are left to the student to investigate. He must clearly 
understand that the problem can be solved only for a given or an 
assumed speed, because of the necessity of using the active layer 
characteristic OX (Fig. 43). 

Prob. 6. In a direct-current machine it is desired to have a full 
load a flux density of not less than 3500 maxwells per sq.cm. at the 
pole-tip at which the commutation takes place; the m.m.f. across the 
active layer is 7500 amp .-turns, the air-gap is 11 mm., the air-gap 
factor 1.25. The ratio of the ideal pole width to the pole pitch is 0.7. 
What are the permissible ampere-turns on the armature, per pole? 
Solution: The net m.m.f. across the active layers at the pole-tip under 
consideration is 3500X0.8X1.1X1.25=3850 amp.-tums. Hence 
Ma-iUC) X0.7t =7500-3850=3650. Ans. i(AC)T=5200. 

Prob. 7. The specific electric loading in a direct-current machine 
is 250 amp.-conductors per centimeter; the average flux density on 
the pole-face is 8.5 kl. per sq.cm., at no. load and at the rated full load 
terminal voltage; the width of the equivalent ideal pole is 32 cm. The 
estimated internal voltage drop at full load is about 5 per cent of the 
terminal voltage. Calculate the ampere-turns per pole required to 
compensate for the transverse armature reaction; the active layer 
characteristic (Fig. 43) is as follows: 

Jlf=4 56 7 8 9 11 13 kilo ampere-turns ; 
B = 5.40 6.75 7.75 8.50 8.90 9.20 9.55 9.78 kl. per sq.cm. 

Ans. 950. • 
Prob. 8. For the preceding machine, calculate the total required 
excitation at full load, per pole; the brush shift is 8 per cent of the 
pole-pitch; the pole-pitch t=46 cm.; 2000 ampere-tums are required 
for the parts of the machine outside the active layer. 

Ans. 12,000 amp.-tums. 
Prob. 9. A shimt-woimd motor is designed so as to operate at a 
certain speed at full load. Show how to predict its speed at no-load. 



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172 THE MAGNETIC CIRCUIT [Art. 54 

Prob. 10- Show how, in a compound-wound motor, the total 
required field excitation must be divided between the shunt and series 
windings in order to obtain a prescribed speed regulation between no- 
load and full load. 

Prob. 11. Assume the curve OX in Fig. 43 to be given in the form 
of an analytic equation, B'^f(M). Show that the unknown excitation 
Oa'='M' is determined by the equation 2M2f(M)=F(M'+M2)- 
F{M'-M^), where the function F is such that dF(M)/dM^f(M); 
3/o = Oa is the excitation corresponding to the given value of 6 or J5 = a6. 

Prob. 12. Apply the formula of the preceding problem to the case 
when the active layer characteristic can be represented by (a) the log- 
arithmic curve y = a log (1 +6x) ; (6) the hyperbola y^gx/{h+x) ; (c) a 
part of the parabola {y^—y^) =2p(x— x©), continued as a tangent straight 
line passing through the origin. 

64. Commutating Poles and Compensating Windings. The 

two limiting factors in proportioning a direct-current machine are, 
first, the sparking imder the brushes, and secondly, the armature 
reaction. In order to reverse a considerable armature current 
in a coil during the short interval of time that the coil is under a 
brush, an external field of a proper direction and magnitude is 
necessary. In ordinary machines (Fig. 42) this field is obtained 
by shifting the brushes so as to bring the short-circuited armature 
conductors under a pole fringe. However, with this method the 
specific electric loading and the armature ampere-tums must be 
kept below a certain limit with reference to the ampere-tums on 
the field ; otherwise the armature reaction would weaken the field 
to such an extent as to reduce the flux density in the fringe below 
the required value. Therefore, in many modem machines, instead 
of moving the brushes to the poles, part of the poles, so to say, are 
brought to the brushes (Fig. 44). These additional poles are 
called commutating poles or interpoles. Their polarity is under- 
stood with reference to Fig. 42 : Since the E brush had to be shifted 
toward the north pole, now a north interpole is placed over each E 
brush. 

The armature belts Te, Te, create a strong m.m.f . along the axis 
of the commutating pole Nc, in the wrong direction. Therefore, 
the winding on each interpole must be provided first with a num- 
ber of ampere-tums equal and opposite to that of the armature, 
and secondly with enough additional ampere-tums to establish 
the required commutating flux. These additional ampere-tums 
are calculated only for the air-gap, armature teeth, and the pole 
body itself. The m.m.f. required for the armature core and the 



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Chap. IX] ARMATURE REACTION IN D.C. MACHINES 



173 



yoke of the machine is negligible, because the commutating flux is 
small as compared with the main flux and is displaced with respect 
to it by ninety electrical degrees. The winding on the interpoles 
is connected in series with the main circuit of the machine, because 
the armature m.m.f. is proportional to the armature current, and 
also because the density of the reversing field must be proportional 
to the current undergoing commutation. 

The flux density under the interpoles is determined from the 
condition that the e.m.f. induced in the armature conductors by 
the commutating flux be equal and opposite to the e.m.f. due to 
the inductance of the armature coils imdergoing commutation. 
For practical purposes, the inductance can be only roughly esti- 
mated (see Art. 68 below), but on the other hand an accurate cal- 

Compensating Commutating 
Main pole Winching /pole or interpole 




W 



T«, Te 

^^ Generator 

« 4mm 



^xxxvycixxx-xxxxx vHT^ 



Motor 



H^ 



Fig. 44. — Interpoles and a compensating winding in a direct-current machine. 

dilation is not necessary, because the number of ampere-turns on 
the interpole is easily adjusted by a shunt around its winding, as 
in the case of a series winding on the main poles. Or else the 
commutating flux can be increased by " shimming up " the 
interpoles. The m.m.f. required for establishing a required 
commutating flux is calculated in the same manner as in the case 
of the main flux, viz., the saturation curves (Figs. 2 and 3) are 
used for the pole body and the teeth, while the reluctance of the 
air-gap is estimated as is explained in Arts. 36 and 37. The 
commutating poles must be of such a width that all the coils 
undergoing commutation are under their influence. 

A comparatively large leakage factor, say between 1.4 and 1.5, 
or over, is usually assumed for the commutating poles, on accoimt 
of the proximity of the main poles. It is advisable to concentrate 
the winding near the tip of the interpole, in order to reduce the 



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174 THE MAGNETIC CIRCUIT [Art. 54 

magnetic leakage. The leakage factor of the main poles is also 
somewhat increased by the presence of the interpoles; this is one 
of their disadvantages. Some other disadvantages are : the ven- 
tilation of the field coils is more difficult, and a smaller ratio of 
pole arc to pole pitch must be used. However, the advantages 
gained by the use of commutating poles are such that their use 
is rapidly becoming imiversal. 

The interpole winding removes the effect of the transverse 
belts T, Tj in the commutating zone, but does not neutralize their 
distorting effect under the main poles. Hence, the distortion and 
its accompanying reduction of the main flux are practically the 
same as without the interpoles. To remove this distortion a cow- 
pensating winding (Fig. 44), connected in series with the main cir- 
cuit of the machine, is sometimes placed on the main poles. The 
connections are such that the compensating winding opposes the 
magnetizing action of the armature winding. By properly select- 
ing the specific electric loading of the compensating winding the 
transverse armature reaction under the poles can be removed, 
either completely or in part. This winding was invented inde- 
pendently by Deri in Europe and by Professor H. J. Ryan in this 
country; on account of its expense, it is used in rare cases 
only. 

When a compensating winding is used in addition to the inter- 
poles, the number of ampere-turns on the interpoles is consider- 
ably reduced, because the compensating winding can be made to 
neutralize the larger portion or all of the armature reaction. In 
such a machine a much higher specific loading can be allowed than 
in an ordinary machine of the same dimensions. Therefore, such 
compensated machines are particularly well adapted for rapidly 
fluctuating loads, and for sudden overloads or reversals of rotation 
in the case of a motor. 

Prob. 13. From the following data determine the ampere-turns 
required on each commutating pole of a turbo-generator: The com- 
mutating poles are made of cast steel; the average flux density on 
the face of the interpole is 6000 maxwells per sq.cm.; the pole-face 
area 250 sq.cm.; the pole cross-section 160 sq.cm.; the radial length 
of the interpole 27 cm.; the leakage factor, 1.5. The air-gap reluctance 
is 2.7 millirels, the true tooth density 20 kilolines per sq.cm., the height 
of the tooth 4.5 cm., and the armature ampere-turns per pole 9500. 

Ans. About 15,300. 

Prob. 14. The rated current of the machine in the preceding problem 



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Chap. IX] ARMATURE REACTION IN D.C. MACHINES 176 

is 1200 amp., and in addition to the interpoles the machine is to be 
provided with a compensating winding. Show that each interpole 
should have at least 8 turns, and each main pole be provided with 10 
bars for the compensating winding, in order to neutralize the armature 
distortion imder the main poles and provide the proper commutating 
field. 

Prob. 16. Machines provided with interpoles are very sensitive 
as to their brush position. By shifting the brushes even slightly from 
the geometrical neutral the terminal voltage of such a generator can be 
varied to a considerable extent. In the case of a motor the speed can 
be regulated by this method, without adjusting the field rheostat. Give 
an explanation of this " compounding" effect of the brush shift. 

66. Armature Reaction in a Rotary Converter. The actual 
currents of irregular form which flow in the armature winding of a 
rotary converter may be considered as the resultants of the direct 
current taken from the machine and of the alternating currents 
taken in by the machine. The resultant magnetizing action upon 
the field is the same as if these two kinds of currents were flowing 
through two separate windings. Therefore, the armature reaction 
in a rotary converter can be calculated by properly combining the 
armature reactions of a synchronous motor and of a direct-current 
generator. 

The alternating-current input into a rotary converter may be 
either at a power factor of imity, if the field excitation is properly 
adjusted, or the input may have a lagging or a leading component, 
the same as in the case of a synchronous motor. The armature 
reaction due to the energy component of the input consists chiefly 
in the distortion of the field, against the direction of rotation of the 
armature. But the action of the direct current is to distort the 
field in the direction of rotation, and since the two m.mis. are not 
much different from one another, the resultant transverse arma- 
ture reaction is very small. The direct reaction of the direct cur- 
rent depends upon the position of the brushes, and the direct 
reaction due to the alternating currents is determined by the 
reactive component of the input, which component may vary 
within wide limits. Thus, the resultant direct reaction of a rotary 
converter may be adjusted to almost any desired value. 

The ohmic drop in the armature of a rotary converter has a 
different expression than in either a direct-current or a synchro- 
nous machine, because the vh- loss must be calculated for the actual 
shape of the superimposed currents. Rotary converters are some- 



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176 THE MAGNETIC CIRCUIT [Art. 55 

times provided with interpoles, in order to improve the commu- 
tation and to use a higher specific electric loading.^ 

^ For further details in regard to the armature reaction in a rotary con- 
verter see Arnold, WechseUtronUechnik, Vol. 4 (1904), Chap. 28; Pichel- 
mayer, Dynamobau (1908), p. 276; Standard Handbook, index imder '' Con- 
verter, synchronous, effective annature resistance ''; Parshall & Hobart^ 
Electric Machine Design (1906) p. 377; Lamme and Newbury, Interpoles in 
Synchronous Converters, Trane. Amer. Inst Electr. Engrs., Vol. 29 (1910), 
p. 1626. 



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CHAPTER X 
ELECTROMAGNETIC ENERGY AND INDUCTANCE 

66. The Energy Stored in an Electromagnetic Field. Experi- 
ment shows that no supply of energy is required to maintain a 
constant magnetic field. The power input into the exciting coil or 
coils is in this case exactly equal to that converted into the t^r heat 
in the conductors, and this power is the same whether a magnetic 
-field is produced or not. Another famiUar example is that of a 
permanent magnet, which maintains a magnetic field without any 
supply of energy from the outside, and apparently without any 
decrease in its internal energy. Nevertheless, every magnetic field 
has a certain amount of energy stored within it, though in a form 
yet imknown. This is proved by the fact that an expenditure of 
energy is required to increase the field, and, on the other hand, 
when the flux is reduced, some energy is returned into the 
exciting electric circuit. 

The conversion of electric into magnetic energy and vice 
versa is accomplished through the e.m.f. induced by the vary- 
ing flux. Let a magnetic flux be excited by a coil CC (Fig. 45) 
supplied with current from a source of constant voltage E, and let 
there be a rheostat r in series with the coil. Let part of the resist- 
ance in the rheostat be suddenly cut out in order to increase the 
current in the coil. It will be found that the current rises to its 
final value not instantly; namely, when the current increases, the 
flux also increases, and in so doing it induces in the electric circuit 
an e.m.f., say e, which tends to oppose the current. This e, 
together with the iR drop, balances the voltage E, so that for a 
time the power Ei supplied by the source is larger than the power 
i^R lost in the total resistance of the circuit. The difference, 
Ei —i?Rj is stored in the magnetic field created by the coil and by 
the other parts of the circuit. The energy eidt supplied by the 
electric circuit during the element of time dt is converted into the 
magnetic energy of the field, by the law of the conservation of 

177 



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178 



THE MAGNETIC CIRCUIT 



[Art. 56 



energy, while the amount {E —e)idt=='i?Rdt is converted into 
heat. 

If now the same resistance is suddenly introduced into 
the circuit, the current gradually returns to its former value, and 
during this variable period the 'PR loss is larger than the power Ei 
supplied by the source. The applied voltage is in this case assisted 
by the voltage e induced by the decreasing field, the e.m.f . e sup- 
plying part of the v^R loss. 

To make the matter more concrete let the source of electric 
energy be a constant-voltage, direct-current generator, driven by 




Fig. 45. — ^The magnetic field produced by a coil, showing complete and 

partial linkages. 

a steam turbine. Let the load consist of coils of variable resist- 
ance Rj and let the coils produce a considerable magnetic field. 
As long as the load current is constant, the rate of the steam con- 
sumption is determined by the i^R loss in the circuit. When the 
current increases, the steam is consumed at each instant at a 
higher rate than it would be with a constant current of the same 
instantaneous value. The energy of steam is thus partly stored 
in the magnetic field of the coils. When the current is returned 
to its former value, the steam consumption during the transitional 



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Chap. X] ENERGY AND INDUCTANCE 179 

period is less than that which corresponds to the "PR loss in the 
circuit, so that practically the same amount of steam is saved 
which was expended before in increasing the magnetic field. 

These phenomena may be explained, or at least expressed in 
more familiar terms, by assuming the magnetic field to be due to 
some kind of motion in a medium possessed of inertia (Art. 3). 
When the field strength is increased it becomes necessary to accel- 
erate the parts in motion, overcoming their inertia. When the 
field is reduced, the kinetic energy of motion is returned to the 
electric circuit. One can also conceive of the energy of the mag- 
netic field to be static and in the form of some elastic stress. 
Under this hypothesis, when a current increases, the magnetic 
stress also increases at the expense of the electric energy. In 
either case, when the current is constant no energy is required 
to maintain the field, any more than to maintain a constant rota- 
tion in a fly-wheel or a constant stress in an elastic body. 

It seems the more probable that the magnetic energy of a 
circuit is stored in some kinetic form, because the current which 
accompanies the flux is itself a kinetic phenomenon. On the other 
hand, it appears more likely that electrostatic energy is due to some 
elastic stresses and displacements in the medium, and thus it 
may be said to be potential energy. Electric oscillations and 
waves consist, then, in periodic transformations of electrostatic into 
electromagnetic energy,. or potential into kinetic energy, and vice 
versa, similar to the mechanical oscillations and waves in an 
elastic body. In the familiar case of current or voltage resonance 
the total energy of the circuit at a certain instant is stored in the 
form of electrostatic energy in the condenser (permittor) con- 
nected into the circuit, or in the natural permittance of the circuit; 
the current and the magnetic energy at this instant are equal 
to zero. At another instant, when the current and the magnetic 
field are at their maximum, the energy stored is all in the form 
of magnetic energy, and the voltage across the condenser and 
the stress in the dielectric are equal to zero. An oscillating 
pendulum offers a close analogy to such a system. The resistance 
of the electric circuit, and the magnetic and dielectric hysteresis, 
are analogous to the friction and windage which accompanies 
the motion of the pendulum. 

As it is of importance in mechanical machine design to know 
the inertia of the moving parts of a machine, so it is often necessary 



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180 THE MAGNETIC CIRCUIT [Art. 57 

in the design and operation of electrical apparatus and circuits to 
know the amount of energy stored in the magnetic field, or the 
electro-magnetic inertia. This inertia modifies the current and 
voltage relations in the electric circuit in somewhat the same way 
in which the inertia of the reciprocating parts in an engine modi- 
fies the useful effort. In mechanical design a revolving part is 
characterized by its moment of inertia from which the stored 
energy can be calculated for any desired speed. So in electrical 
engineering a circuit or an apparatus is characterized by its electro- 
magnetic inertia or inductance j from which it is possible to calculate 
the magnetic energy stored in it at any desired value of the cur- 
rent. In this and in the two jiext chapters expressions are deduced 
for the inductance of some of the principal types of apparatus 
used in electrical engineering. 

Prob. 1. A stationary electromagnet attracts and lifts its armature 
with a weight attached to it. Explain how the energy necessary for 
the lifting of the weight is supplied by the electric circuit. 

Prob. 2. A direct-current generator driven by a water-wheel is 
subjected to very large and sudden fluctuations of the load, which 
the governor and the gate mechanism are unable to follow properly. 
A heavy fly-wheel on the generator shaft would improve the operating 
conditions. Would a reactance coil in series with the main circuit 
achieve the same result, provided that it could be made large enough 
to store the required excess of energy ? 

Prob. 3. What experimental evidence could be offered to support 
the contention that the energy of an electric circuit is contained in the 
magnetic field linked with the current, and not in the current itself ? 
The flow of current is usually compared to that of water in a pipe; 
is not all the kinetic energy stored in the moving water itself and not 
in the surrounding mediimi, and if so, is a current of electricity really 
like a current of water ? 

Prob. 4. Describe in detail current and voltage resonance * and 
free electrical oscillations, from the point of view of the periodic conversion 
of electromagnetic into electrostatic energy and vice versa, taking account 
of the dissipation of energy in the resistance of the circuit. 

67. Electromagnetic Energy Expressed through the Linkages 
of Current and Flux. In order to obtain a general expression for 
the energy stored in the magnetic field of an electric circuit, con- 
sider first a single loop of wire aa (Fig. 11) through which a steady 
electric current i is flowing. Let the cross-section of the wire be 
small as compared to the dimensions of the loop, so that the flux 

* See the author's Experimental Electrical Engineering, Vol. 2, pp. 17 to 25. 



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Chap. X] ENERGY AND INDUCTANCE 181 

inside the wire may be disregarded. The electromagnetic energy 
possessed by the loop is equal to the electrical energy spent in 
establishing the current i in the loop against the induced e.m.f . 
Let it and ^t be the instantaneous values of the current in 
amperes and the flux in webers at a moment t dining the period of 
building up the flux, and let et be the instantaneous applied 
voltage. Let the flux increase by d0t during an infinitesimal 
interval dt; then the electrical energy (in joules) supplied from 
the source of power to the magnetic field is 

dW -=it€tdt =it (d0t/dt)dt =itd0t, 

where— et= —d0t/dt is the instantaneous e.m.f. induced in the 
loop by the changing flux. The total energy supplied from the 
electrical source during the period of building up the field to its 
final value is 



W 



= Citd0t (98) 

•^0 



Li a medium of constant permeability the integration can be 
easily performed, because the flux is proportional to the current, 
or, according to eq. (2) in Art. 5, 0t =(Pitj where (P is the permeance 
of the magnetic circuit, in henrys. Eliminating by means of this 
relation either it or 9t from eq. (98) we can obtain any one of the 
following three expressions for the electromagnetic energy stored 
in the loop: 

W=ii^(P; (99) 

In the first form, eq. (99) expresses the fact that the magnetic 
energy stored in a loop is equal to one-half the product of the cur- 
rent by the flux; in the second form, it shows that the stored 
energy is proportional to the square of the current and to the per- 
meance of the magnetic circuit. Both forms are of importance in 
practical applications. 

Take now the more general case of a coil of n turns (Fig. 45) ; 
the flux which links with a part of the turns is now of a magnitude 
comparable with that of the flux which links with all the turns of 
the coil. We shall consider the complete linkages and the partial 
linkages separately. Consider first the energy due to the flux 



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182 THE MAGNETIC CIRCUIT [Art. 57 

which links with all the turns of the coil. The e.m.f. induced in 
the coil by this flux, when the current changes, is — e« = — n(d(P</ctt), 
and the relation between the current and the flux is 0t=^c^iu 
where (Pc is the permeance of the path of the complete linkages. 
By repeating the reasoning given above in the case of a single loop 
we find that 

W,=ini(P,, (100) 

or 

Wc=inh^(Pc, (100a) 

where the subcript c signifies that the quantities refer to the com- 
plete linkages only (Fig. 45). Two forms only are retained, being 
those that are of the most practical importance. 

The energy of the partial linkages is calculated in a similar 
manner. Let J(Pp be a small annular flux (Fig. 45) which links 
with Up turns of the coil, where rip may be an integer or a fraction. 
For these turns the linkage with J(^p is a complete linkage, while 
for the external (n—rip) turns it is no linkage at all and represents 
no energy, because no e.m.f. is induced by J(Pp in the turns external 
to it. Thus, the energy due to the flux J^p, according to eqs. 
(100) and (100a), is equal to inpiJ^p, or to iup^Jcpp, The 
total energy of the partial linkages is the sum of such expressions, 
over the whole flux (^p, or 

Wp=ii^npJ0p, (101) 

or 

Fp=it22np2J(Pp (101a) 

The total energy of the coil is 

TF=ii[n(^e + SnpJ(^p], (102) 

or 

F=it>2(P^ + 2np2J(Pp], .... (102a) 

where the first term on t^e right-hand side refers to the complete 
linkages and the second to the partial linkages of the flux and the 
current. In these expressions the current is in amperes, the fluxes 
in webers, the permeances in henrys, and the energy in j|ules 
(watt-seconds) . If other units are used the corresponding numeri- 
cal conversion factors must be introduced. 



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Chap. X] ENERGY AND INDUCTANCE 183 

Some new light is thrown upon these relations by using the 
m.m.f . M instead of the ampere-tums ni. Namely, eqs. (102) and 
(102a) become: 

TF=MM,*e + SMpJ*p], .... (103) 
or 

TF=i[Me2tP,+SMp2j(Pp]. . . . (103a) 

These expressions are analogous to those for the energy stored in 
an electrostatic circuit, viz., ^EQ, and ^E^C (see The Electric Cir- 
cuit). The m.m.f. Mc is analogous to the e.m.f. E] the magnetic 
flux 9c is analogous to the electrostatic flux Q, and the permeance 
(Pc is analogous to the permittance C, 

We can assume as a fimdamental law of nature the fact that 
with a given steady current the magnetic field is distributed in such 
a way that the total electromagnetic energy of the system is a 
maximum. All known fields obey this law, and, in addition, it can 
be proved by the higher mathematics. Eq. (102a) shows that this 
law is fulfilled when the sum n^CPc + lln^(Pp is a maximum. When 
the partial linkages are comparatively small, the energy stored is a 
maximum when the permeance (Pc of the paths of the total linkages 
is a maximum. This fact is made use of in the graphical method 
of mapping out a magnetic field, in Art. 41 above. 

Prob. 5. The no-load saturation curve of an 8-pole electric gen- 
erator is a straight line such that when the useful flux is 10 megalines 
per pole the excitation is 7200 amp .-turns per pole; the leakage factor 
is 1.2. Show that at this excitation there is enough energy stored 
in the field to supply a small incandescent lamp with power for a few 
minutes. 

Prob. 6- Explain the fimction and the diagram of connections of 
a field-discharge switch. 

Prob. 7. Prove that the magnetic energy stored in an apparatus 
containing iron is proportional to the area between the saturation curve 
and the axis of ordinates. The saturation curve is understood to give 
the total flux plotted against the exciting ampere-turns as abscissse. 
Hint: See Art. 16. 

Prob. 8. Deduce expression (102a) directly, by writing down an 
expression for the total instantaneous e.m.f. induced in a coil (Fig. 45). 

Prob. 9. Explain the reason for which, in the formulae deduced above, 
it is permissible to consider n to be a fractional number. 

68. Inductance as the CoeflSicient of Stored Energy, or the 
Electrical Inertia of a Circuit. Eq. (102a) shows that in a mag- 
netic circuit of constant permeability the stored energy is propor- 



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184 THE MAGNETIC CIRCUIT [Art. 68 

tional to the square of the current which excites the field. The 
coefficient of proportionality, which depends only upon the form 
of the circuit and the position of the exciting m.m.f., is defined 
as the inductance of the electric circuit. The older name for 
inductance is the coefficient of self-induction. It is assumed 
here that the magnetic circuit is excited by only one electric cir- 
cuit, so that there is no mutual inductance. Thus, by definition 

W=iLP (104) 

where the inductance is 

L=ri2(P,+Srip2J(Pp, (105) 

or, replacing the summation by an integration, 

L^n^(Pc+C\p^d(Pp (106) 

Since the permeances in eq. (102a) are expressed in henrys, and 
the numbers of turns are numerics, the inductance L in the defin- 
ing eqs. (105), or (106), is also in henrys. If the permeances aTe 
measured in millihenrys or in perms, the inductance L is measured 
in the same imits. As a matter of fact, the henry was originally 
adopted as a imit of inductance, and only later on was applied to 
permeance.^ 

In some cases it is convenient to replace the actual coil (Fig. 45) 
by a fictitious coil of an equal inductance, and of the same number of 
turns, but without partial linkages. Let (Peq be the permeance of 
the complete linkages of this fictitious coil; then, by definition, 
eqs. (105) and (106) become 

L=n^(Peq (106a) 

This expression is used when the permeance of the paths is calcu- 
lated from the results of experimental measurements of inductance, 
because in this case it is not possible to separate the partial 
linkages. Use is made of formula (106a) in chapter XII, in cal- 
culating the inductance of armature windings. 

* The use of the henry as a unit of permeance was proposed by Professor 
Giorgi. See Trans. Intern. Elec. Congress at St. Louis (1904), Vol. 1, p. 136. 
The connection between inductance and permeance seems to have been first 
established by Oliver Heaviside; see his Electromagnetic Theory (1894), Vol. 
1, p. 31. 



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Chap. X] ENERGY AND INDUCTANCE 185 

The inductance L is related in a simple manner to the electro- 
motive force induced in the exciting electric circuit when the cur- 
rent varies in it. Namely, the electric power supplied to the cir- 
cuit or returned from the circuit to the source is equal to the rate 
of change of the stored energy, so that we have from eq. (104) 

dW/dt =i( -e) ^U(cli/dt), 
or, canceling i, 

e = -L{di/dt), (107) 

The sign minus is used because e is understood to be the induced 
e.m.f . and not that applied at the terminals of the circuit. There- 
fore, when dW/dt is positive, that is, when the stored energy 
increases with the time, e is induced in the direction opposite to that 
of the flow of the current, and hence by convention is considered 
negative. Inductance is sometimes defined by eq. (107), and then 
eqs. (104) and (105) are deduced from it. The definition of L by 
the expression for the electromagnetic energy seems to be a more 
logical one for the purpose of this treatise, while the other defini- 
tion in terms of the induced e.m.f. is proper from the point of 
view of the electric circuit. 

Looking upon the stored magnetic energy as due to some kind 
of a motion in the medium, eq. (104) suggests the familiar expres- 
sions ^mv^ and ^KaJ^ for the kinetic energy of a mechanical system.^^ 
Taking the curre nt to be analogous to the velocity of motion , the \ 
inductancebecomes an alogous to mechanical mass an d moment of 
inertia. The larger the electromagnetic inertia L the more energy 
is stored with the same current. Equation (107) also has its^ 
analogue in mechanics, namely in the familiar expressions mdv/dt 
and Kd(o/dt for the accelerating force and torque respectively. 
The e.m.f. e represents the reaction of the circuit upon the source 
of power when the latter tends to increase i the rate of flow of elec- 
tricity. While these analogies should not be carried too far, they 
are helpful in forming a clearer picture of the electromagnetic 
phenomena. 

The role of inductance, L, in the current and voltage relations of 
alternating-current circuits is treated in detail in the author's 
Electric Circuit. In this book inductance is considered from the 
point of view of the magnetic circuit, i.e., as expressed by eqs. (104) 
to (106) . In the next two chapters the values of inductance are 



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186 THE MAGNETIC CIRCUIT [Art. 58 

calculated for some important practical cases, from the forms and 
the dimensions of the magnetic circuits, using the fimdamental 
equations (104), (105) and (106). The reader will see that the 
problem is reduced to the determination of various permeances and 
fluxes; hence, it presents the same difficulties with which he is 
already familiar from the study of Chapters V and VI. 

Inductance of electric circuits in the presence of iron. When 
iron is present in the magnetic circuit, three cases may be distin- 
guished: 

(1) The reluctance of the iron parts is negligible as compared 
to that of the rest of the circuit; 

(2) The reluctance of the iron parts is constant within the 
range of the flux densities used ; 

(3) The reluctance of the iron parts is considerable, and is 
variable. 

In the first two cases, eqs. (104), (105) and (106) hold true, and 
the inductance can be calculated from the constant permeances of 
the magnetic circuit. In the third case, inductance, if used at all, 
must be separately defined, because eq. (1026) does not hold when 
the permeance of the circuit varies with the current. The equa- 
tion of energy is in this case 

W ^nCit^^tc-^^CnpiM^^tv). . . (108) 

This equation is deduced by the same reasoning as eq. (98). 

The following three definitions of inductance are used by differ- 
ent authors when the reluctance of a magnetic circuit is variable : 
(a) the expressions (104) and (108) are equated to each other, and 
L is calculated separately for each final value i. of the current. 
Thus L is variable, and neither eq. (105) nor (107) hold true. (6) 
L is defined from eq. (107) ; in this case neither eq. (104) nor (105) 
are fulfilled, (c) L is defined at a given current by eq. (105) so 
that Li represents the sum of the linkages of the flux and the cur- 
rent. Therefore eq. (107) becomes e == —d(Li)/dt, and dW =id(Li). 
With each of the three definitions L is variable, and therefore is not 
very useful in applications. The author's opinion is that when the 
permeance of the circuit is variable, L should not be introduced at 
all, but the original equation of energy (108) be used directly. Or 
else in approximate calculations, a constant value of L can be 
used, calculated for some average value of i or 0. 



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Chap. X] ENERGY AND INDUCTANCE 187 

Prob. 10. It is desired to make a standard of inductance of one 
millihenry by winding uniformly one layer of thin flat conductor upon 
a toroidal wooden ring of circular cross-section. How many turns are 
needed if the diameter of the cross-section of the ring is 10 cm. and 
the mean diameter of the ring itself is 50 cm.? Ans. 400. 

Prob. 11. An iron ring of circular cross-section is uniformly woimd 
with n turns of wire, the total thickness of the winding being t; the 
mean diameter of the ring is D and the radius of its cross-section r. 
What is the inductance of the apparatus, assiuning the permeability 
of the iron to be constant and equal to 1500 times that of the air? 
Hint: d(Pp = fi2n{r+x)dx/7cD; np^n(x/t), 

Ans. 1.25(nVZ>) [1500r»+^(^+iO]XlO-» henry. 

Prob. 12. A ring is made of non-magnetic material, and has a 

rectangular cross-section of dimensions h and h) the mean diameter 

of the ring is D. It is uniformly wound with n timis of wire, the total 

thickness of the winding being t. What is the inductance of the winding? 

Ans. 0.4(nVZ>) [bA +f ^(6 +h +30J10-* millihenry. 

Prob. 13. The ring in the preceding problem has the following 
dimensions: D=50 cm.; A=6«10 cm.; it is to be wound with a con- 
ductor 3 mm. thick (insulated). How many turns are required in order 
to get an inductance of 0.43 of a henry? Hint : Solve by trials, assum- 
ing reasonable values for t\ the number of turns per layer decreases as 
the thickness of the winding increases. Ans. About 5300. 

Prob. 14. It is desired to design a choke coil which will cause a 
reactive drop of 250 volts, at 10 amp. and 50 cycles. The cross-section 
of the core (Fig. 12) is 120 sq.cm., and the mean length of the path 
130 cm.; the maximum flux density in the iron must be not over 7 
kl. per sq. cm. What is the required number of turns and the length 
of the air-gap in the core?* Ans. 150; 3.7 mm. 

Prob. 16. An electrical circuit, which consists of a Leyden jar 
battery of 0.01 mf. capacity and of a coil having an inductance of 10 
millihenrys, imdergoes free electrical oscillations in such a way that the 
maximum instantaneous voltage across the. condenser is 10,(X)0 volts. 
What is the current through the inductance one-quarter of a cycle 
later, neglecting any loss of energy during the interval ? 

Ans. 10 amp. 

Prob. 16. Suggest a practical experiment which would prove directly 
that the stored electromagnetic energy is proportional to the square 
of the current. 

Prob.' 17. Prove that the inductance of a coil of given external 
dimensions is proportional to the square of the number of turns, taking 
into account the complete and the partial linkages. Show that the 
ohmic resistance of the coil is also proportional to the square of the 
number of turns, provided that the space factor is constant. 

Note 1. The theoretical calculation of the inductance of short 

* For a complete design of a reactive coil see G. Kapp, Transformers (1908), 
p. 106. 



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188 THE MAGNETIC CIRCUIT [Art. 58 

straight coils and loops of wire in the air is rather complicated, because 
of the mathematical difficulties in expressing the permeances of the 
paths. Those interested in the subject will find ample information 
in Rosa and Cohen's Formuloe and Tables for the Calculation of Mutual 
and Sdf-Inductancey in the Bulletins of the Bureau of Standards, Vol. 
5 (1908), No. 1. The article contains also quite a complete bibliography 
on the subject. See also Orlich, Kapazitdt und Inductivitiit (1909), 
p. 74 et seq. 

Note 2. In the formulae deduced in this and in the two following 
chapters, it is presupposed that the current is distributed uniformly 
over the cross-section of the conductors. Such is the case in conductors 
of moderate size and at ordinary commercial frequencies, imless perchance 
the material is itself a miagnetic substance. With high frequencies, 
or with conductors of unusually large transverse dimensions, as also with 
conductors of a magnetic material, the current is not distributed uni- 
formly over the cross-section of the conductors, the current density 
being higher near the periphery. The result is that, as the frequency 
increases, the inductance becomes lower and the ohmic resistance higher. 
This is known as the skin effect. For an explanation, for a mathematical 
treatment in a simple case, and for references see Heinke, Handbuch 
der Elektrotechnik, Vol. 1 (1904) part 2, pp. 120 to 129. Tables and for- 
mulae will be found in the Standard Handbookj and in Foster's Pocket- 
Book, See also C. P. Steinmetz, AUernating-Currevd Phenomena (1908), 
pp. 206-208, and his Transient Electric Phenomena (1909), Section III, 
Chapter .VII; Arnold, Die Wechselstromtechnik, Vol. 1 (1910), p. 564; A. 
B. Field, Eddy Currents in Large Slot-wound Conductors, Trans. Amer, 
Inst. Electr. Engrs., Vol. 24 (1905), p. 761. 



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CHAPTER XI. 

THE INDUCTANCE OF CABLES AND OF TRANS- 
MISSION LINES. 



Let a direct current of 



69. The Inductance of a Single-phase Concentric Cable. Let 

Fig. 46 represent the cross-section of a concentric cable, which 
consists of an inner core A and of an external annular conductor 
£>, with some insulation C between them. Let the radii of the 
conductors be a, 6, and c, respectively. The insulation outside of 
D and the sheathing are not shown, 
amperes flow through the inner 
conductor away from the reader 
and return through the outer 
conductor. The magnetic field 
produced by this current links 
with the current, and for reasons 
of symmetry the lines of force 
are concentric circles. The field 
is confined within the cable, 
because outside the external con- 
ductor D the m.m.f . is i —i =0. 

In the space between the two 
conductors the lines of force are 
linked with the whole current, 
and since there is but one turn, 
the m.m.f. is equal to i. The length of a line of force of a radius x 
cm. is 27ra:sothat the magnetic intensity isff =i/2;rx, amp. turns per 
cm., the corresponding flux density B =;ii/2;rx maxwells per sq.cm. 
Thus, the flux density decreases inversely as the distance from the 
center; it is represented by the ordinates of the part ^ of a hyper- 
bola. 

In the space within the inner conductor A, a line of force of 
radius x is linked with a current i, =i{nx^/naP) =i(x/a)2, provided 

189 




Fig. 46— -The magnetic field within 
a concentric cable. 



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190 THE MAGNETIC CIRCUIT [Art. 59 

that the current is distributed uniformly over the cross-section of 
the conductor. The length of the line of force is 2nx, so that 
H =^iJ2nx=xi/{2na^)y and B =^ jxxi/ (^na?) . Thus, in this part 
of the field the flux density increases as the distance from the 
center and is represented by the straight line Oq. 

In the space inside the conductor D, a line of force of a 
radius a: is linked with the current —i{x^-'V^)/((?—ly^) of the 
external conductor and with the current +i of the internal con- 
ductor, or altogether with the current ig=^i{c^ —x^)/ (c^ —b^). 
Consequently, here, the flux density, is represented by the 
hyperbola rs, the equation of which is 

B^tdJ2nx-^ld{c^/x -x)/[2;r(c2 -IP)], 

The curve Oqra gives a clear physical picture of the field dis- 
tribution in the cable, and helps one to imderstand the linkages 
which enter into the calculation of the inductance of the cable. 

The inductance of the cable is calculated according to the 
fundamental formula (106), the complete linkages being in the 
space between the two conductors, and the partial linkages being 
within the space occupied by the conductors themselves. Con- 
sider a piece of the cable one centimeter long. The permeance of a 
tube of force of a radius x and of a thickness dx is ijdx/2nx, so that 
the permeance of the complete linkages is, 

Lc' = (Pc = r iidx/2nx = {ii/2n)lsi(b/a) perm/cm., (109) 

where Ln is the symbol for natural logarithms. In this case the 
permeance is equal to the inductance because the number of turns 
n = l. The sign "prime" indicates that the quantities L/ and 
(Pc refer to a imit length of the cable. 

For the space inside the inner conductor rip = (x/a)^, this being 
the fraction of the current with which the line of force of radius x 
is linked. Hence, the part of the inductance of the cable due to 
the field inside the conductor A is 

L/ = (fi/2n) C{x/a)Hdx/x) =fi/S7r =0.05 perm/cm. (110) 

This formula shows that the part of the inductance due to the 
field within the inner conductor is independent of the radius of the 



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Chap. XI] INDUCTANCE OF TRANSMISSION LINES 191 

conductor, and is always equal to 0.05 perm, per cm., or 0.05 milli- 
henry per kilometer length of the cable. 

The exact expression for the part of the inductance of the cable 
Ljy' due to the linkage within the outer conductor is given in 
problem 10 below. The formula is rather complicated for prac- 
tical use, especially in view of the fact that this part of the induct- 
ance is comparatively small, because the flux density on the part 
rs of the curve is small. It is more convenient, therefore, to make 
simplifying assumptions, when the thickness t of the outer con- 
ductor is small as compared to 6. Namely, the length of all the 
paths within the outer conductor may be assumed to be equal 
to 2;r6, so that the permeance of an infinitesimal path of a radius x 
and thickness dx nearly equals iidx/2nb. Furthermore, the volume 
of the current in the outer conductor, between the radii 6 and x 
may be assumed to be proportional to the distance a:— 6, and 
hence equal to i (x —b)/(c — 6) . A line of force of a radius x is linked 
therefore with the whole current i in the inner conductor and with 
the above-stated part of the current in the outer conductor, and, 
since the currents flow in opposite directions, this line of 
force is linked altogether with the current i(c —x)/{c —b). Hence, 
it is linked with rip = (c — x)/(c — 6) turns. Thus,the inductance of 
the cable, due to the outer partial linkages, is, in the first approxi- 
mation, 

L^; =fi/{27:bfi) C\c -x)Hx =^t/b perm/cm. . (Ill) 

If a closer approximation is desired, it is convenient to expand 
eq. (114) in ascending powers of t/b, as is explained in problem 11. 
The result is 

W =T^t/b[l -^(t/b)^+^(t/b)^ . . .] perm/cm. . (112) 

It will thus be seen that eq. (HI) is an accurate approximation, 
because eq. (112) contains in the parentheses no term with the 
first power of the ratio t/b. 

Thus, the total inductance of a concentric cable, I kilometers 
long, is 

L =[0.46 Logio (b/a) 4-0.05+L^']Z miUihenrys, . (113) 

where Lj/ is given by eqs. (Ill), (112), or (114), according to the 
accuracy desired. Expressions (110) and (114) are correct only at 
low frequencies, such as are used for power transmission. With 



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192 THE MAGNETIC CIRCUIT [Art. 59 

very high frequencies, the skin effect becomes noticeable, that is, 
the current in the inner conductor is forced outward and that in 
the outer conductor inward. In the limit, when the frequency 
is infinite, the currents are concentrated on the opposing surfaces 
of the conductors, and the partial linkages are equal to zero. 
Thus, each of the expressions in question must be multiplied by a 
variable coefficient k which, for a given cable, is a function of 
the frequency. At ordinary frequencies *; = 1, and gradually 
approaches zero as the frequency increases to infinity.^ 

Prob. 1. A concentric cable is to be designed for 750 amperes, the 
current density to be about 2.2 amp. per gross sq.mm., and the thickness 
of the insulation between the conductors to be 6 mm. What are the 
dimensions of the conductors assuming them to be solid, that is, not 
stranded? The fact that they are in reality stranded is taken care of 
in the permissible current density. 

Ans. a « 10.5 ; h « 16.5 ; c = 19.6 mm. 

Prob. 2. Plot the curve Oqra of distribution of the flux density 
in the cable given in the preceding problem. 

Ans. At x=* a, B^ 143; at x =6, B^ 91 maxwells per sq.cm. 

Prob. 3. What is the total flux in megalines per kilometer of the 
cable specified in the two preceding problems? Ans. 15.1. 

Prob. 4. Show how to plot the curve of the distribution of flux density 
in a three-phase concentric cable, at some given instantaneous values 
of the three currents. 

Prob. 6. What is the inductance of a 25-km. cable in which the 
diameter of the inner conductor is 12 mm., the thickness of insulation 
is 3 mm., and the dimension c is such that the current density in the 
outer conductor is 10 per cent higher than in the inner one ? 

Ans. 25 [0.0810 + 0.050 + 0.0122] = 3.58 millihenry. 

Prob. 6. A cable consists of three concentric cylinders of negligible 
thickness; the radii of the cylinders are ri, r2, and rj, beginning with 
the inner one. What is the inductance in millihenrys per kilometer, 
when a current flows through the inner cylinder and returns equally 
divided through the two others? 

Ans. 0.46 [log (rt/ri) + 0.25 log {r^/rt) .] 

Prob. 7. In the cable given in the preceding problem the total 
current i flows through the middle cylinder, the part mi returns throi^h 
the inner cylinder, and the rest, m, returns through the outer one. What 
is the total inductance per kilometer of length? 

Ans. 0.46 [m* log (fj/rO +n* log (rs/r^)]. 

* For the field distribution in and the inductance of non-concentric cables 
see Alex. Russell, AUemating Currents, Vol. 1 (1904), Chap. XV; for the 
reactance of armored cables see J. B. Whitehead, '\The Resistance and React- 
ance of Armored Cables, Trans. Amer. Inst. Electr. Engra., Vol. 28 (1909), 
p. 737. 



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Chap. XI] INDUCTANCE OF TRANSMISSION LINES 193 

Prob. 8. At what ratio of 6 to a in Fig. 46 is the magnetic energy 
stored within the inner conductor equal to that stored between the 
two conductors ? Ans. 1.28. 

Prob. 9. It is required to replace the solid inner conductor A in 
Fig. 46 by an infinitely thin shell of such a radius a' that the total 
inductance of the cable shall remain the same. What is the radius of 
the shell? Hmt: (ju/27r)Ln(a/a')='j"/8w. 

Ans. a7a = £-'-''^ =0.779. 

Prob. 10. Prove that the part of the inductance due to the linkages 
within the outer conductor in Fig. 46 is expressed by 

Lb' - 2nJ^h')^ '''^''^'''^^ ""*^^''' ^^' -4c'6')]. . . (114) 

Hint: d(9p^tidx/2nx\ np^l-n(x*-h^)Mc*/h^. 

Prob. 11. Deduce eq. (112) from formula (114), assuming the 
ratio t/b to be small as compared to unity. Hint: Put c = 6(1+2/) 
where y^t/h is a small fraction. Expand Ln(l—y) into an infinite 
series, and omit in the nmnerator of eq. (114) all the terms above t/*; 
expand (c^—b^) in the denominator in ascending powers of y, and divide 
the numerator by this polynomial. 

60. The Magnetic Field Created by a Loop of Two Parallel. 
Wires. Let Fig. 47 represent the cross-section of a single-phase 
or direct-current transmission line, the wires being denoted by 
A and B, With the directions of the currents in the wires shown 
by the dot and the cross, the magnetic field has the directions 
shown by the arrow-heads, one-half of the flux linking with each 
wire. Before calculating the inductance of the loop it is instruct- 
ive to get a clear picture, quantitative as well as qualitative, of the 
field itself. 

The field distribution is symmetrical with respect to the line 
AB and the axis 00\ The whole flux passes in the space between 
the wires, so as to be linked with the m.m.f. which produces it, and 
then extends to infinity on all sides. The flux density is at its 
maximum near the wires and gradually decreases toward 00' 
and toward ± oo, as is shown by the curve pqsts'q'p\ The ordi- 
nates of this curve represent the flux densities at the various points 
of the line passing through the centers of the wires. The reason 
for which the flux density is larger near the wires is that the path 
there is shorter, although the m.m.f. acting along all the paths 
is the same. This m.m.f. is numerically equal to the current i in 
the wires, the number of turns being equal to one. 

It is proved below that the magnetic paths outside the conduc- 



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194 



THE MAGNETIC CIRCUIT 



[Art. 60 



tors themselves are eccentric circles, with their centers on the 
line AB extended. The equipotential surfaces are circular 
cylinders, which are shown in Fig. 47 as circles passing through the 
centers A and B of the conductors. Within the conductors them- 
selves there are no equipotential surfaces. 

For purposes of analysis it is convenient to regard the field in 
Fig. 47 as the result of the superposition of two simpler fields, simi- 
lar to those of the concentric cable of the preceding article. Con- 




FiG. 47. — ^The magnetic field produced by a single-phase transmission line. 

sider the conductor A, together with a concentric cylinder of an 
infinitely large radius, as one conducting system. Let the current 
flow through A toward the reader, and return through the infinite 
cylinder. Let the conductor B with a similar concentric cylinder 
form another independent system. The currents in the conduc- 
tors A and B are to be the same as the actual currents flowing 
through them, but each infinite cylinder is to serve as a return for 
the corresponding conductor, as if there were no electrical connec- 
tion between A and B, The currents in the two cylinders are 
flowing in opposite directions and the cylinders themselves 



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Chap. XI] INDUCTANCE OF TRANSMISSION LINES 195 

coincide at infinity, because the distance AB between their axes is 
infinitely small as compared to their radii. Hence, the two cur- 
rents in the cylinders cancel each other, and the combination of 
the two component systems is magnetically identical with the 
given loop. 

In a medium of constant permeability, the resultant magnetic 
intensity H, produced at a point by the combined action of several 
independent m.m.fs., is equal to the geometric sum of the intensi- 
ties produced at the same point by the separate m.m.fs.^ This 
being true of the intensities, the component flux densities at any 
point are also combined according to the parallelogram law 
because they are proportional to the intensities. Hence, the 
resultant flux can be regarded as the result of the superposition 
of the fluxes created by the component systems. 

The field produced by the system A consists of concentric cir- 
cles, the flux density outside the conductor A being inversely pro- 
portional to the distance from the center of A (curve qr in Fig. 46). 
The field created by the system B consists of similar circles around 
B, and the field shown in Fig. 47 is a superposition of these two 
fields. Thus the resultant field intensity jff at a point P is a 
geometric sum of 

Hi=i/27rri, (115) 

and 

H2=i/27rr2, (116) 

ff 1 and H2 being perpendicular to the corresponding radii vectors 
ri and r2 from the centers of the conductors to the point P. The 
directions of Hi and H2 are determined by the right-hand screw 
rule. Since Hi and H2 are known in magnitude and direction at 
each point of the field, the resultant intensity H may also be deter- 
mined. 

To deduce the equation of the lines of force in the resultant 
field, we shall express analytically the condition that the total flux 
which crosses the surface CP is equal to zero, provided that C and 
P lie on the same line of force.'* This total flux may be considered 

* This principle of auperposUion can be considered (a) as an experimental 
fact; (6) as an immediate consequence of the fact that in a medium of constant 
permeability the efifects are proportional to the causes; (c) as a consequence 
of Laplace's law dJET^ Const. X ids sin ^/lOr', according to which the total 
field intensity is regarded as the sum of those produced by the infinitesimal 
elements of the current, or currents. 



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196 THE MAGNETIC CIRCUIT [Abt. 60 

as the resultant of the fluxes due to the systems A and B. Accord- 
ing to eq. (115), the flux density due to A, at a distance x from A, 
is Bi =id/2nXj so that the flux due to A which crosses CP is 

(J>i'= r *Sidx = (/ri/2;r)Ln(ri/i4C)maxwells/cm. . (117) 

•/AC 

This flux is directed to the left, looking from C to P. By analogy, 
the flux due to the system B is 

<^2' = (/n/2;r)Ln(r2/5C)maxwells/cm., . . . (118) 

and is directed to the right, looking from C to P. The condition 
that no flux crosses CP is, that (J>i' is equal to ^2', or 

Ln(ri/.lC)=Ln(r2/BC), 
or 

ri/r2=ilC/5C=Const (119) 

This is the equation of a line of force in " bipolar " co-ordinates; 
the curve is such that the ratio of ri to r2 remains constant, How- 
ever, this constant is different for each line of force, because each 
line has its own point C. 

Eq. (119) may be proved to represent a circle, by selecting an 
origin, say at A, and substituting for ri and r2 their values in terms 
of the rectangular co-ordinates x and y. The following proof by 
elementary geometry leads to the same result. Produce AP and , 
lay off PD=PB=r2. According to eq. (119), BD is parallel to 
CP, and consequently PC bisects the angle APB=(o. Let the 
point C lie on the same line of force with C; then no flux passes 
through PC, and by analogy with eq. (119) we have 

ri/r2=i4C75C'=Const (120) 

By plotting PD^ =r2 (not shown in figure) along PA, in the oppo- 
site direction from PD, and connecting £>' to B, one can show as 
before that PC bisects the angle 5PD =180° -c^. But the 
bisectors of two supplementary angles are perpendicular to each 
other; consequently, CPC is a right angle, and the point P lies 
on a semicircle erected on the diameter CC\ This semicircle is the 
line of force itself, because all the points, such as P, which are deter- 
mined by C and C must lie on it. Another semicircle below the 
line AB closes the line of force. 



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Chap. XI] INDUCTANCE OF TRANSMISSION LINES 197 

From eqs. (119) and (120) the following expressions are obtained 
for the radius R of the line of force : 

BC 

^^i^iBc/Acy ^^2^> 

or 

BC 
^^l+BC'/AC' ^^^^^ 

so that the line of force can be easily drawn for a given C or C. 

To prove that the equipotential lines are also circles we proceed 
as follows. The line AB is evidently an equipotential line, because 
it is perpendicular to all the lines of force. The difference of mag- 
netic potental or the m.m.f. between AB and P, contributed by 
the system A, is equal to i{d\/27t) ampere-tums, where the angle 
BAP is denoted by Oi. This is because the m.m.f. due to the sys- 
tem A, taken around a complete circle concentric with A, is equal 
to i, and is distributed uniformly along the circle, for reasons of 
symmetry. Or else it follows directly from eq. (115). By 
analogy, the difference of magnetic potential between AB and P, 
due to the system B, is i{62/2n). Thus the total difference of 
potential between AB and P is 

^cp=(V2;r)(5i+52)=(i/27r)(;r-a>). . . (123) 

This shows that the m.m.f. between any two points in the field is 
proportional to the difference in the angles oj at which the line AB 
is visible from these points. For any two points on the same 
equipotential line M^ is the same, so that the equation of such a 
line is 

w=const (124) 

This represents the arc of a circle passing through A and B, and 
corresponding to the inscribed angle w. 

Prob. 12. A single-phase transmission line consists of two conductors 
1 cm. in diameter, and spaced 100 cm. between the centers. Draw 
the curve of flux density distribution (pqs/ in Fig. 47) for an instantaneous 
value of the current equal to 100 amp. 

Ans. a;=50.0 25.0 0.5 -0.5 -50.0 -oo cm.; 

B^ 1.60 2.13 80.40 -79.60 -0.53 maxw./sq. cm. 
Prob. 13. For the transmission line in problem 12 draw the lines 



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198 THE MAGNETIC CIRCUIT [Art. 61 

of force which divide the total flux between the wires into ten equal 
parts (not counting the flux within the wires). 

Ans. The circles nearest to 00' cross AB at a distance of 48.4 
cm. from each other. 
Prob. 14. Referring to the two preceding problems, draw ten 
equipotential circles which divide the whole m.m.f. of 100 ampere-turns 
into ten equal parts. 

Ans. The arcs nearest to AB intersect 00* at a distance of 
32.5 cm. from each other. 
Prob. 16. A telephone line runs parallel to a single-phase power 
transmission line. The position of one of the telephone wires is fixed; 
show how to determine the position of the other wire so as to have a 
minimum of inductive disturbance in the telephone circuit. Hint: 
The center lines of the two telephone wires must intersect the same line 
of force due to the power line. 

Prob. 16. A telephone line runs parallel to a 25-cycle, single-phase 
transmission line. The distances from one of the telephone wires to 
the power wires are 3.5 m. and 2.7 m.; the distances from the other 
telephone wire to the power wires are 3.6 and 2.5 m. (in the same order). 
What voltage is induced in the telephone line per kilometer of its length, 
when the current in the power line is 1(X) effective amperes? 

Note: In practice, this voltage is neutralized by transposing 
either line after a certain number of spans. Ans. 0.33 volt. 

61. The Inductance of a Single-phase Line. The inductance of 
a single-phase line (Fig. 47) can be calculated according to the fun- 
damental formulae (105) or (106), provided that the permeances 
of the elementary paths be expressed analytically. However, in 
this case it is much simpler to use the principle of superposition 
employed in the preceding article, and to consider the actual flux 
as the resultant of two fluxes each surrounding concentrically one 
of the wires and extending to inflnity. The fluxes produced by 
the two component systems are equal and symmetrical with 
respect to the wires. It is therefore sufficient to calculate the 
linkages of the loop AB with the flux produced by one of the sys- 
tems, say that corresponding to A, and to multiply the result by 
two. 

The flux produced by A and having il as a center link, with 
the current in the loop AB. These linkages may be divided into 
the following: 

(a) Linkages within the wire A ; that is, from a; =0 to x =a; 

(6) Linkages between the wires A and jB, that is, from 

x=atox=6— a; 



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Chap. XIJ INDUCTANCE OF TRANSMISSION LINES 199 

(c) Linkages outside the loop, that is, from a:=6-fa to x = 
infinity, 

{d) Linkages within the wire 5, that is, from 

a:=6— atox=6+a. 

The linkages (a), (6) and (c) are the same as in a concentric 
cable (Fig. 46), because the shape of the lines of force and the 
number of turns with which they are linked are the same. The 
partial linkages (d) are somewhat difficult to express analytically. 
When the distance b between the wires is large as compared to 
their diameters, the whole current in B may be assumed to be con- 
centrated along the axis of the wire B, instead of being spread over 
the cross-section. With this assumption, the partial linkages {d) 
are done away with, the linkages (6) are extended to a: =6, and the 
linkages (c) begin from x=h. The expressions for the linkages (a) 
and (6) are given by eqs. (110) and (109) respectively. The link- 
ages (c) are equal to zero, because in this region the lines of force 
produced by A are linked with both A and B, and therefore 
with i— i=0 ampere -turns. Thus, the inductance in question is 

L' =0.46 logio(6/a) +0.05 (125) 

This gives the inductance of a single-phase line in perms per centi- 
meter length, or in millihenrys per kilometer length of the wire.^ 
To obtain the inductance per unit length of the line this expression 
must be multiplied by two, because the linkages due to the flux 
produced by the system B are not taken into accoimt in eq. (125). 
However, for the purposes of the next two articles it is more con- 
venient to use expression (125), and to consider separately the 
inductance of each wire, remembering that the two wires of a loop 
are in series, and that therefore their inductances are added .^ 

Prob. 17. Check by means of formula (125) some of the values for 
the inductance and reactance of transmission lines tabulated in the 
various pocketbooks and handbooks. 

* It is of interest to note that the exact integration over the partial linkages 
(d) leads to the same Eq. (125), so that this formula is correct even when 
the wires are close to each other. See A. Russell, Alternating Currents, Vol. 
1 (1904), pp. 59-60. 

' The inductance of two or more parallel cylinders of any cross-section can 
be expressed through the so-called " geometric mean distance," introduced 
by Maxwell. For details see Orlich, Kapasntdt und Induktivitat (1909), 
pp. 63-74. 



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200 THE MAGNETIC CIRCUIT [Art. 61 

Prob. 18. Show by means of tables or curves that the inductance 
of a transmission line varies much more slowly than (a) the spacing 
with a given size of wire, (6) the size of wire with a given spacing. 

Prob. 19. When the diameters of the two wires A and B are different, 
prove that the inductance of the complete loop is the same as if the 
diameter of each wire was equal to the geometric mean of the actual 
diameters. 

Prob. 20. Show that the inductance of a single-phase line with a 
ground return can be calculated from eq. (125) by putting h^2h where 
h is the elevation of the wire above the ground. Hint: In Fig. 47 the 
plane 00' may be considered to be the surface of the earth, assumed 
to be a perfect conductor. If the earth be removed, an "image" con- 
ductor B must be added in order to provide a return path for the 
current, such that the field surrounding A would remain the same. 

Prob. 21. Two single-phase lines are placed on two cross-arms of 
the same tower, one directly above the other, at a vertical distance 
of c cm apart. What is the total inductance of the combination, when 
the two lines are connected in parallel and each line carries one-half 
of the total current? 

Solution : Consider the four wires as forming four fictitious systems, 
with cylinders at infinity as returns. Let h be the spacing in each 
loop, and let h be larger than c. Denote the wires in one loop by 1 
and 2, in the other by V and 2\ and let d be the diagonal distance 
between 1 and 2'. Assume all the wires except 1 to be of an infinitesimal 
cross-section. Then, the linkages of the flux produced by the system 
1 with the currents in the four wires are 

i»L/-0.05(W+0.2(it)> Ln (c/a) +0.2i (ii)Ln(6/c) 

+0.2(ii)' Ln(d/6) miUijoules/km. 

Thus, allowing the same amount for the linkages due to the cur- 
rent in the wire 1', we get that the inductance of the split line, each way, 
is 

L' = 2L/ «i[0.46 logio {hd/ca) +0.05] millihenrys/km., 

instead of the expression (125) for the single line. The same result 
is obtained when h is smaller than c. Hence, by splitting a line in 
two the inductance is considerably reduced, because partial linkages are 
substituted for some of the complete linkages. If d were equal to c 
the reduction would be 50 per cent; but since d is always larger than 
c the gain is less than 50 per cent. However, when the two lines are 
very far apart the saving is very nearly 50 per cent. 

Prob. 22. A certain single-phase transmission line has been designed 
to consist of No. 000 B. & S. conductor with a spacing of 180 cm. It 
is desired to reduce the reactive drop by about 20 per cent, without 
increasing the weight of copper, by using two lines in parallel, with 
the same spacing. What is the size of the conductor and the distance 
between the loops? Ans. No. 1 B. & S. ; about 8 cm. 

Prob. 23. Solve problem 21 when the load is divided \meq\ially 



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Chap. XI] INDUCTANCE OF TRANSMISSION LINES 201 

between the two loops, the currents being mi and ni respectively, where 
m+w — 1. 

Ans. L'-0.46[(7n»+n*) log (c/a) +log {h/c)+2mn log (d/b)] + 

0.05(7n'+n2), when 5^^^^ ^^(1 L'=0.46[(m*+n*) log (6 /a) + 

27nnlog (d/c)]+0.05(m^ -\-n^), when 6<c. 
Prob. 24. A single-phase line consists of three conductors, the 
total current flowing through conductor 1, and returning throi^h con- 
ductors 2 and 3 in parallel. If the current in one return conductor is 
mi, and in the other m, where 7n+n = l, proVe that the inductance of 
the line per kilometer of its length is, in millihenrys, 

L'-0.46 [log (&i2/ai) +m' log (h^s/ch) +n* log (^js/as) +m log (b^/ftis) 

+m iog (bizAs) +n log (bw/bas)] +0.05(1 +m^ +n*), when 5i2> 5is> 623. 

In the particular case when &i2=&28=^i8> 01=02 = 03, and m=n = i, the 
inductance is reduced by 25 per cent as compared to that of a single loop.^ 

62. The Inductance of a Three-phase Line with Symmetrical 
and Semi-symmetrical Spacing. The magnetic field which sur- 
rounds a single-phase line varies in its intensity from instant to 
instant, as the current changes, but the direction of the magnetic 
intensity and of the flux density at each point remains the same. 
In other words, the flux is a pulsating one. The field created by 
three-phase currents in a transmission line varies at each point in 
both its magnitude and direction. At the end of each cycle, the 
field assumes its original magnitude and direction. If the spacing 
of the wires is symmetrical, the field at the end of each third of a 
cycle has the same magnitude and position with respect the next 
wire. The field may therefore be said to be revolving in space. 

This revolving flux, like that in an induction motor, induces 
e.m.fs. in the three phases. The problem is to determine these 
counter-e.m.fs. in the transmission line, knowing the size of the 
wires, the spacing, and the load. In transmission line calculations, 
especially in determining the voltage drop and regulation, it is 
convenient to consider each wire separately, and to determine the 
voltage drop in phase and in quadrature with the current. Thus, 
having expressed the e.m.fs. induced by the revolving flux in terms 
of the constants of the line, each wire is then considered as if it 
were brought outside the inductive action of the two other wires. 

We shall consider first the case of an equidistant spacing of the 
three wires, because in most practical calculations of voltage drop 

^The splitting of conductors discussed in problems 21 to 24 has been 
proposed for extremely long transmission lines, in order to reduce their induct- 
ance and at the same time increase their electrostatic permittance (capacity). 



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202 THE MAGNETIC CIRCUIT [Art. 62 

an iinsymmetrical spacing is replaced by an equivalent equidistant 
spacing. The exact solution for an unsymmetrical spacing is given 
in the next article. Let the instantaneous values of the three cur- 
rents in the wires A, B and C of a three-phase transmission line be 
ii, i2 and 13. The sum of the three currents at each instant is 
zero, or 

ii -1-12 +13=0 (126) 

Let *etf be the equivalent flux which links at any instant with the 
wire A. The instantaneous e.m.f. induced in this wire is 

ei^-d^eq/dt (127) 

The equivalent flux consists of the actual flux outside the wire plus 
the sum of the fluxes inside the wire, each infinitesimal tube of flues 
being reduced in the proper ratio, according to the fraction of the 
cross-flection of the wire with which it is linked. Or, what is the 
same thing, each wire is replaced by an equivalent hollow cylinder 
of infinitesimal thickness, without partial hnkages, as in problem 
9 in Art. 59 (consult also the definition of equivalent permeance 
given in Art. 58). 

In order to determine ^eq we replace the three-phase system 
by two superimposed single-phase systems. The current ii in the 
wire A may be thought of as the sum of the currents —12 and —is, 
each flowing in a separate fictitious wire, and both of these wires 
coinciding with A. The currents -h 12 in 5 and —12 in A form 
one single-phase loop, while the currents -his in C and —is in A 
form the other loop. The flux 0eq which surrounds A is the sum 
of the fluxes produced by these two loops. The flux per imit 
length of the line, due to the first loop, is equal to —(Pgq%, since 
the number of turns is equal to one. For the same reason (P'^ = 
U where U is determined by eq. (125). Hence, the flux per 
unit length of the line, due to the first loop, is — L'i2. Similarly, 
the flux due to the second loop and linked with A is equal to 
— L'is, the same value of U being used because the spacing and 
the sizes of all of the wires are the same. Thus, 

0eq=-L'i2-L'is=LHi, .... (128) 

the last result being obtained by substituting the value of t2-ht3 
from eq. (126). Thus, eq. (127) becomes simply 

ei^-Udii/dt, (129) 



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Chap. XI] INDUCTANCE OF TRANSMISSION LINES 203 

that is, the induced e.m.f . is the same as in a single-phase line carry- 
ing the current ii. Thus, the inductance of a three-phase linemth 
symmetrical spacing, per wire, is the same as the inductance of a 
single-pha^e line, per wire, with the same size of wire and the same 
spacing. The total e.m.f. induced in each wire is in quadrature 
with the current in the wire. 

In reaching this conclusion the following facts were made use 
of : (a) The current in each wire at any instant is equal to the sum 
of the currents in the two other wires; (6) the fluxes due to sepa- 
rate m.m.fs. can be superimposed in a medium of constant perme- 
ability; (c) The inductance of the loop A-B is equal to that of A-C 
because of the same spacing. No other suppositions in regard to 
the character of the load or the voltages between the wires were 
made. Therefore, the conclusion arrhred at holds true : 

(a) For balanced as well as unbalanced loads; 

(6) For balanced or unbalanced line voltages; 

(c) For a three-wire two-phase system, three-wire single- 
phase system, monocyclic system, etc. 

(d) For sinusoidal voltages as well as for those departing from 
this form. 

Semi-symmetrical Spacing. When two out of the three distances 
between the wires in a three-phase line are equal to each other, the 
arrangement is called semi-symmetrical. Two common cases of 
this kind are : (a) When the wires are placed at the vertices of an 
isosceles triangle; (6) when they are placed at equal distances in 
the same plane, for instance on the same cross-arm, or are fastened 
to suspension insulators, one above the other. In such cases the 
inductive drop in the symmetrically situated wire is the same as if 
the wire belonged to a single-phase loop, carrying the same current, 
and with a spacing equal to the distance of this wire to either of 
the other two wires. Let, for instance, the distance A-B be equal 
to B^, and let the distance A-^ be different from the two. The 
proof given above can be repeated for the wire B, and the same 
conclusion will be reached because the spacing A-C is not used 
in the deduction. But, of course, the proof does not hold true for 
either wire A or C. 

When the three wires are in the same plane, the inductance of 
each of the outside wires is larger than that of the middle wire. 
This can be shown as follows : Let the three wires be in a horizontal 
plane, and let them be denoted from left to right by A, B, C. Let 



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204 THE MAGNETIC CIRCUIT [Art. 62 

the distance between A and B be equal to h and the distance 
between A and C be equal to 26. If the wire B were moved to 
coincide with C, the inductance of A would be the same as if it 
belonged to a single-phase loop with a spacing 26. If C were 
moved to coincide with -B, the inductance of A would be that of a 
wire in a single-phase loop with a spacing 6. Thus, the inductance 
of A corresponds in reality to a spacing intermediate between 6 and 
26. The inductance of the middle wire B is the same as that of a 
wire in a single-phase loop with the spacing 6, as is proved above. 
Thus, the inductance of either il or C is larger than that of B. 

An inspection of a table of the inductances or reactances of 
transmission lines will show that the inductance increases much 
more slowly than the spacing. For instance, according to the 
Standard Handbook , the reactance per mile of No. 0000 wire, at 
25 cycles, is 0.303 ohm with a spacing of 72 inch, and is 0.340 ohm 
with a spacing of 150 inch. Therefore, in practical calculations, 
when the spacing is semi-symmetrical, the values of inductance are 
taken the same for all the three wires, for an average spacing 
between the three, or, in order to be on the safe side, for the maxi- 
mum spacing. In the most unfavorable case, even if an error of 
say 10 per cent be made in the estimated value of the inductance, 
and if the inductive drop is say 20 per cent of the load voltage, the 
error in the calculated value of voltage drop is only 2 per cent of 
the load voltage, and that at zero power factor. At power factors 
nearer imity, when the vector of the inductive drop is added at 
an angle to the line voltage, the error is much smaller. 

Prob. 26. Show that the instantaneous electromagnetic energy 
stored per kilometer of a three-phase line with symmetrical spacing is 
equal to iL'(ii*+t2^+t8*) millijoules per kilometer, where U has the 
value given by eq. (125). If this is true, then each wire may be con- 
sidered as if it were subjected to no inductive action from the other 
wires and had an inductance U expressed by eq. (125). This is another 
way of proving eq. (129), and the statement printed in italics above. 
Solution: Consider each wire, with a concentric cylinder at infinity, 
as a component system. Determine the linkages of the field created 
by the system A with the currents in A, B, and C, as in Art. 61. The 
result is equal to \Uii^, Similar expressions are then written by analogy 
for the fluxes due to the systems B and C 

Prob. 26. Show graphically that, when the distances A-B and 
A-C are equal, the equivalent flux linking with A is independent of 
the spacing B-C, and is the same as if J5 and C coincided. That is, 
prove that the inductance of A is the same as if it belonged to a single- 



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Chap. XI] INDUCTANCE OF TRANSMISSION LINES 



205 



phase loop. Solution: Let /i, /,, and /a (Fig. 48), represent the vectors 
of the three currents at an unbalanced load. The current /i in A is 
replaced by —I2 and — /s, and the system is split into two single-phase 
loops, A-B and A-€, The fluxes due to these systems and linked with 
A are denoted by <l^i2 and 0^. They are in phase with the corresponding 
currents, and are proportional to the magnitudes of these currents, 
becavse of the equal spacing. Hence, the triangles of the currents and 
of the fluxes are similar, and the resultant flux 0i linking with A is 
in phase with /i. If <l^i2 = i(P/a, and i<l>i=i(P/8, where (P is the equivalent 
permeance bleach single-phase loop, then the result shows that #i = i(P/i. 
If the wires B and C coincided the equivalent permeance (P would be 
the same, and hence the proposition is proved. The voltage drop, Ei, 
due to the flux 0^ is shown by a vector leading /i by 90 degrees. 

63. The Equivalent Reactance and Resistance of a Three- 
phase Line virith an Unequal Spacing of the Wires. In the case 
of an unequal spacing of the wires eqs. (126) and (127) still hold 
true, because they do not depend upon the spacing; but eq. (128) 
becomes 

^eq=-L\2i2-'L'isis, (130) 

where U12 is the value of the inductance per unit length, calcu- 
lated by eq. (125) for the spacing between A and 5, and L'13 is 
the value of the inductance per unit 
length, for the spacing 4-C Substi- ii 

tuting the valueof 0eq from eq. (130) 
into eq. (127) we get 

ei =L'i2dt2/^ + L'lsdis/dt. (131) 

This shows that with an unequal 
spacing the effect of the mutual 
induction of the phases cannot be 
replaced by an equivalent inductance 
in each phase, because, generally 
speaking, the currents 1*2 and i^ cannot 
be eliminated from this equation by 
means of eq. (126). 

Let us apply now eq. (131) to 
the case of sinusoidal currents and 

voltages. Let the current in the wire B be t2= ^2/2 sin 2nftf 
where 1 2 is the effective value of the current ; then the first term 
on the right-hand side of eq. (131) becomes 27r/L'i2V2Z2 cos 2nft. 




Fig. 48. — ^The currents and 
fluxes in a three-phase line 
with a symmetrical spacing. 



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206 THE MAGNETIC CIRCUIT [Art. 63 

In the symbolic notation this is represented as jx\2l, where a;x2' 
=2nfLi2 is the reactance corresponding to Li2f I2 is the vector 
of the effective value of the current in the wire 5, and ; signifies 
that the vector X12I2 is in leading quadrature with the vector I2, 
Consequently, the voltage drop Ei in the wire A, equal and 
opposite to the induced e.m.f ., is 

Ei = -jxi2'l2-ixxz'I^ (132) 

When the currents are given, I2 and Iz can be expressed in the 
usual way through their components, and the drop E^ is then 
expressed through its components as ei-\-je\. The reactances 
X12' and xiz are taken from the available tables, for the specified 
frequency and the appropriate spacings, or else they can be 
calculated using the value of U from eq. (125). 

The voltage drop Ei in eq. (132) can be represented as if it were 
due to an equivalent reactance x\ and an equivalent resistance ri 
in the phase A (the latter in addition to the actual ohmic resistance 
of the wire). This is possible when 1 2 and Iz can be expressed in 
terms of /i, and is especially convenient whenever the phase differ- 
ence between these currents and their ratio is constant. Namely, 
let the current I2 lead the current l\ in phase by ^12 electrical 
degrees (Fig. 49) . Then 

f2 = (/2//i)fi(cos^i2+/sin^i2), • . . (133) 

where (J2/I1) is the ratio of the effective values of the currents, 
apart from their phase relation. Multiplyingtthe vector 7i by 
{h/h) changes its magnitude to that of I2, while multiplying it by 
(cos <^i2 +/ sin <l>i2) turns it counter-clockwise by ^12 degrees. By 
analogy we also have that 

/3 = (/3//i)/i(cos<^i3-l-;sin^i3). . . . (134) 

Both ^12 and <f)iz are measured counter-clockwise. Substituting 
these values into eq. (132) and separating the real from the 
imaginary part we get 

Ei=Ii[{l2/Ii)^i2 sin 4>i2+{h/Ii)xiz sin ^13] 

''iIl[(l2/Il)Xi2' C0B<l>i2+(Iz/Il)XlZ COQ^lzl . . (135) 

Thus, the drop Ei is the same as if it were caused by a fictitious 
reactance 

Xi = - {l2/h)Xl2 cos ^12 - (/3//l)a:i3' cos ^13, . (136) 



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Chap. XIJ INDUCTANCE OF TRANSMISSION LINES 



207 



and a fictitious resistance 

n' = {I2/Ii)xi2' sin ^12 + {h/Ii)xi3' sin ^13. • (137) 

Both Xi' and ri may be either positive or negative, depending 
upon the constants of the circuit and of the load. The resistance 
ri' does not involve any loss of 
power, converted into heat; it 
merely shows that energy is trans- 
ferred inductively from phase A 
into B or C, at a rate Ir^i, due to 
a Jack of symmetry in the resultant 
field. 

Prob. 27. Show that with a 
balanced three-phase load 



r/- 0.866 (X; 






(138) 




Prob. 28. When the three wires 
are In the same plane, the spacings 
being equal, and the three-phase load 
balanced, show that the equivalent reactance of each outside wire 



Fig. 49.— The currents and fluxes 
in a three-phase line with an 
unsymmetrical spacing. 



a;o'-a^'+0.435/XlO-3 ohm/km.. 



(139) 



where Xm' is the reactance of the middle wire per kilometer, m ohms. 
The equivalent resistance of the middle wire is zero, and that of the 
two outside wires is 

To' = ±0.753/X 10-3 ohm/km., (140) 

where the sign pliLS refers to the wire in which the ciurent leads that in 
the middle wire. 

Prob. 29. Compare the vector diagram in Fig. 49 with that in Fig. 
48, and shown that with an unsymmetrical spacing the induced voltage 
El is not in quadrature with the corresponding current /i, so that the 
action of the other two wires cannot be replaced by an equivalent 
inductance alone, but only by an inductance and a resistance. Show 
graphically that the latter may be either positive or negative. 



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CHAPTER Xn 

THE INDUCTANCE OF THE WINDINGS OF 
ELECTRICAL MACHINERY. 

64. The Inductance of Transformer Windings. When a 
transformer is operated at no load, i.e., with its secondary cir- 
cuit open, practically the whole flux is concentrated within 
the iron core. When, however, the transformer is loaded, so that 
considerable currents flow in both windings, appreciable leakage 
fluxes are formed (Fig. 50), which are linked partly with the 
primary winding, and partly with the secondary winding. When 
the load current is considerable, the primary and the secondary 
ampere-turns are large as compared to the exciting ampere- 
turns, so that at each instant the secondary ampere-turns are 
practically equal and opposite to the primary ampere-turns. 
An inspection of Fig. 50 will show that the m.m.f. acting upon 
the useful path in the iron is equal to the difference between the 
m.m.fs. of the primary and secondary windings, while the m.m.f. 
acting upon leakage paths is equal to the sum of the m.m.fs. 
of both windings. 

Take, for instance, the line of force fghk; with respect to the 
part fg of its path, the secondary coil Si and the adjacent half 
Pi of the primary coil form together a fictitious annular coil 
(leakage coil). The m.m.f. of this coil is equal to injii, where 
ii is the primary current, and Ui is the number of turns in the 
whole primary coil P. Similarly, the coil ^2 ai^d the part P2 
of the primary coil may be said to form another fictitious coil 
linking with the part hk of the path of the lines of force. 

It will be seen from the dots and crosses that the m.m.fs. 
of the two fictitious coils assist each other, and that the paths 
of the leakage flux are as indicated by the arrow-heads. Some 
lines of force are linked with the total m.m.f. of the fictitious 
coils, others are linked with only part of the turns. Although 

208 



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Chap. XII] 



INDUCTANCE OF WINDINGS 



209 



the reluctance of the leakage paths is very high as compared 
to that of the useful path in iron, yet the m.m.f. acting upon 



Jkctual Field 



Sftsplified Flold 




Fiux Deiialti'- Diatiril>atioa 



X 

s, 

X 




P. 


• 

i 

• 


• 

Pi 

• 


J' 


X 

Si 
"X 


*Mbf 


t-Or* 


<- 




h-* 


*ar* 


♦iflsr 



Fig. 60. — ^The leakage field in a transformer with cylindrical coils. 

the leakage paths is also many times greater than that acting 
upon the path in iron. As a consequence, the leakage fluxes 
reach appreciable magnitudes. 



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210 THE MAGNETIC CIRCUIT [Art. 64 

The leakage fluxes induce e.m.fs. in the windings in lagging 
quadrature with the respective currents, and thus affect the 
voltage regulation of the transformer. That part of the applied 
voltage which balances these e.m.fs. is known as the reactance 
drop in the transformer. It is customary to speak about the 
primary reactance and the secondary reactance, also about the 
primary and the secondary leakage fluxes, as if they had a 
separate and independent existence. However, it must be under- 
stood that each leakage flux is produced by the combined action 
of both windings, as is explained above. Moreover, where the 
leakage fluxes enter the iron they cpmbine with the main flux 
in the proper direction, so that they form there only a com- 
ponent of the resultant flux. 

In reality, the primary ampere-turns are not exactly equal 
and opposite to the secondary ampere-turns, so that, in addition 
to the leakage fluxes shown in Fig. 50, there is a leakage flux 
due to the magnetizing ampere-turns. However, this correction 
is negligible, when the load is considerable, and the calculation 
of the leakage flux is greatly simplified by assuming the primary 
ampere-turns to be exactly equal and opposite to the secondary 
ampere-turns. 

The effect of the leakage reactance upon the performance 
of a transformer is treated in The Electric Circuit; there the 
value of the reactance is supposed to be given. Here the problem 
is to show how to calculate the leakage inductance from the given 
dimensions of a transformer. The two types of winding to be 
considered are the one with cylindrical coils (Fig. 60) and the 
one with flat coils (Fig. 51). Both types of winding can be used 
with any of the three kinds of magnetic circuit which are used 
with transformers (Figs. 12, 13, and 14). 

The problem of calculating the leakage inductance, according 
to the fundamental formula (106), is reduced to that of finding 
the permeances of the individual paths of the leakage flux. It 
would be out of the question here to determine the actual paths 
and to express their permeances mathematically. Therefore, in 
accordance with Dr. Kapp's proposal,^ simplified paths are 
assumed, shown in Fig. 50 to the right. The inductance so 
calculated is corrected by an empirical factor, obtained from 
experiments on transformers of similar type and proportions. 

» G. Kapp, Tranaformera (1908), p. 177. 



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Chap. XII] INDUCTANCE OF WINDINGS 211 

The simplifying assumptions are (a) that the paths within and 
between the coils are straight lines, and (6) that the reluctance 
of the paths in the space outside the coils is negligible, because 
the cross-section of these paths is practically unlimited. 

(1) Cylindrical Coils. We shall calculate first the primary 
inductance of a transformer having cylindrical coils, i.e., the 
inductance due to the linkages of the leakage flux with the 
primary winding. The permeance of the path of the complete 
linkages is (Pci = fi.cLiOm/2l perms, where 0^ is the mean length 
of a turn in the coil P, and ai is the radial thickness of the flux. 
The notation is shown in the detail drawing, at the bottom of Fig. 
50. In this expression aiOm is the mean cross-section of the 
path, being an average between the cross-sections within the 
spaces Pi-zSi and P2-S2. The length of the paths within' the 
coils is 2Z, and the reluctance of the paths outside the coils is 
neglected. This path is linked with nj turns. 

Similarly, the permeance of an infinitesimal annular path 
within the primary coil, at a distance x from its center, and 
of a width dx, is d(Ppi = fiOmdx/2l perms. This path is linked 
with npi = ni(2x/bi) turns. Substituting these values into eq. 
(106) we obtain 

Li = {fmi^0J2l) [a, +£^'\2x/b,)Mx] , 

or 

Li = (/mi2OV20(ai + i6i) perms (141) 

By a somewhat similar reasoning we should find for the com- 
bination of the two secondary coils, assuming them to be connected 
electrically in series, 

L2={fm2^0„,/2I)(a2+ib2) perms (142) 

In the operation of a transformer it is the total equivalent 
inductance of the two windings reduced to one of the circuits 
that is of importance. Since resistances and reactances can be 
transferred from the primary circuit to the secondary or vice 
versa, when multiplied by the square of the ratio of the numbers 
of turns (Art. 446), the equivalent inductance, reduced to the 
primary circuit, and per leg of the core, is 

L^=Li + (ni/n2)2L2 

^k(fini^0m/2l) [a+i(6i+62)]10-8 henrys. . (143) 



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212 THE MAGNETIC CIRCUIT [Art. 64 

All the lengths here are expressed in centimeters, and /x= 1.257, 
also a=ai +02 is the spacing between the coils, which is a known 
quantity. Thus, the unknown distances ai and 02 which enter 
into the expressions for Lj and L2 are eliminated from the 
formula for the equivalent inductance. 

The coefficient k corrects for the difference between the 
actual linkages shown in Fig. 50 at the left, and the assumed 
linkages shown at the right. The values of k, found from experi- 
ments, vary within quite wide limits, depending upon the pro- 
portions of the coils. For good modem transformers Arnold 
gives the limits of k between 0.95 and 1.05.1 See also eq. (147) 
below. Formula (143) gives the inductance of one leg only; 
the equivalent inductance of the whole transformer depends 
upon the electrical connections between the coils. 

In designing a transformer the coils are usually arranged 
in such a way as to reduce the leakage reactance to the least 
possible amount.2 Eq. (143) shows that in order to achieve 
this result, a comparatively small number of turns must be used, 
and the coils must be thin and long. The space a between the 
coils must be kept as small as is compatible with the require- 
ments for insulation and cooling. 

The usual arrangement of coils shown in Fig. 50 gives a 
considerably smaller leakage inductance than the simpler arrange- 
ment shown in Fig. 12. Namely, with the arrangement shown 
in Fig. 12, the permeance of the path of the complete linkages 
in the space between the coils is fiaiOm/l- This expression 
differs from that used before in that I stands in the denominator 
in place of 21. For the partial linkages np=ni(x/bi)j where 
X is measured now from the edge of the primary coil, furthest 
from the secondary coil. Thus, the primary inductance is in 
this case 

or 

Li = (iemi2O„/0(ai+i6i). 

By symmetry we can write the expression for L2, and hence, 

*E. Arnold Wechselstromtechnik (2d edition), Vol. 1., p. 661. 
^ In some cases a considerable leakage reactance is specified as a protection 
against violent short circuits. 



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Chap. XII] INDUCTANCE OF WINDINGS 213 

after combining, 

Lea=H^i^0m/l)[a-\-i(bi+b2)]l(r^y . . . (144) 

This value is between two and four times as large as the 
value given by eq. (143). For this reason, in most transformers, 
the low-tension coil is split into two sections; compare also with 
Fig. 14. 

Formula (143) and the values of k given above have been 
deduced for the core-type transformer. It is clear, however, 
that the same formulae will apply to the shell-type and the cruci- 
form type transformers with cylindrical coils, though the coeffi- 
cient k may have different values in each case. Until more 
reliable and numerous experimental data are available the same 
values of k will have to be used for these types as for the core- 
type.i 

(2) Flat Coils. With flat coils (Fig. 51) the inductance of 
a part of the winding between AB and CD can be calculated 
in precisely the same way as in Fig. 50. If the primary winding 
is split into q sections, the inductance per section, by analogy 
with eq. (143), is 

Le^/q=k[ii(ni/qyOm/2l][a + i(h + b2)]10-»henTy8, . (145) 

where the dimension I is again measured in the direction of 
the lines of force and Om is the mean length of a turn. The 
dimensions a, 6i, and 62 are indicated in Fig. 51. The inductance 
of the whole winding is 

L^^k(fmi'Orn/2ql)[a+l(bi + b2)]10-^hejiTya, . . (146) 

where all the lengths are expressed in centimeters, and //= 1.257. 

This formula presupposes that the m.m.fs. are balanced, or 
in other words, that there are two half-sections of the same 
winding at the ends; such is usually the case in order to reduce 
the leakage reactance. (See also Fig. 13.) 

Eq. (146) shows that the leakage reactance is considerably 
reduced, and consequently the voltage regulation improved, by 
subdividing the windings and placing the primary and the 

* See also the Standard Handbook for Electrical Engineers under Trans- 
former, leakage reactance. 



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214 



THE MAGNETIC CIRCUIT 



[Art. 64 



secondary windings on the core alternately. At a given voltage, 
and with a given type of construction, the spacing a between 
the coils may be considered as constant and independent of the 
number of sections. In transformers for extra-high voltages a 
is large as compared to J (61 +62)7 so that the leakage reactance 
is almost inversely proportional to the number of sections q. 
In low-voltage transformers a is small as compared to 61 and 
62; hence, L^q is almost inversely proportional to ^, because 
61 and 62 are themselves inversely proportional to q. Thus, 




Fig. 61. — ^The leakage field in a transformer with flat coils. 

in general, the inductance of a transformer is inversely pro- 
portional to q^, where n has a value between 1 and 2.i 

Dr. W. Rogowski has given an exact mathematical solution 
for the flux distribution in the case of flat transformer coils, 
assuming the coils and the core to be indefinitely long in the 
direction perpendicular to the cross-section shown in Fig. 51. ^ 

* For experimental data in regard to the effect of the subdivision and 
arrangement of transfonner windings upon the leakage reactance see Dr. W. 
Rogowski, MiUeilungen Ueher Forschungsarheiten, Heft 71 (Springer, 1909), 
p. 18, also his article Ueber die Streuung des Transf ormators, Elektrotechnische 
Zeitschrift, Vol. 31 (1910), pp. 1035 and 1069; also Faccioli, Reactance of 
Shell-type Transformers, Electrical World, Vol. 55 (1910), p. 941. 

2 Dr. W. Rogowski, loc, cU. 



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Chap. XII] INDUCTANCE OF WINDINGS 216 

With certain simplifying assumptions he arrived at the same 
formula for inductance as eq. (146) in which approximately 

fc = 1 - (61 + 62 + 2a) /(2;rZ) . 

Because of the simplifying assumptions made in the deduc- 
tion of this formula, the values of k calculated from the results 
of tests on actual transformers differ slightly from those given 
by the formula. Let k' be an empirical correction coefficient, 
then 

k=k'[l-{bi+b2+2a)/(2nl)] (147) 

In Dr. Rogowski's experiments the actually measured induct- 
ance was on the average 6 per cent higher than the calculated 
one. Until more experimental data are available it is therefore 
advisable to use in eq. (147) the value of A' =1.06. Eq. (147) 
is applicable to transformers of all the three types (Figs. 12 to 
14), though in the case of a shell-type or cruciform transformer, 
the presence of iron outside the coils tends to increase the value 
of k\ However, Dr. Rogowski states, that with the space usually 
allowed for insulation between the coils and the iron, the influence 
of the iron in increasing the leakage reactance is negligible. Eq. 
(147) holds approximately true for cylindrical coils also, though 
there are as yet no conclusive tests for the value of the cor- 
rection factor to be used with such coils. 

The general similarity between the equations for leakage induct- 
ance given above raises the question, as to what element they 
possess in common. This is found in the conception of a leakage 
coil, which is the " fictitious coil " spoken of above. An inspec- 
tion of Figs. 50 and 51 will show that the successive lines of 
force converge upon lines which may be called the " hearts ^' 
of the flux system. These hearts are located in places where 
the net m.m.f. is zero. This is at the edge of the half-coils and 
at the center of the whole-coils, in the two figures mentioned. 
In deriving eq. (144) for the case where the coils are not split, 
the heart is assumed to be at the edge of both coils. If we 
define a leakage coil as that part of the winding between two 
successive hearts, then eq. (144) will always apply to it. In 
eq. (143) 61 and 63 refer to the width of the double leakage coil, 
hence if we substitute in eq. (144) ibi and ^62 for 61 and 62, 



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216 THE MAGNETIC CIRCUIT [Art. 64 

we get the coefficient J, which appears in eq. (143). In eq. 
(143) rii and Leq refer to the double leakage coil. Substituting 
in eq. (144) Jni for Ui and JLeg for Le,, we get eq. (143). Thus, 
the differences between the two equations is readily explained. 
In case the coils are divided in any unusual manner, we must 
first locate the hearts by noticing where the m.m.fs. are balanced. 
Then we should figure out the inductance by eq. (144) for each 
leakage coil separately. The only precaution to be observed is 
that the various quantities refer to the leakage coil. Finally 
(if the coils are in series) we should add the various induct- 
ances together. The arrangement with half coils on the ends 
gives the minimum of inductance for a given number of coils. 

Prob. 1. The approximate assumed dimensions of a 1 5-kw., 2200/1 1 0- 
v., 60-cycle, cruciform-type transformer with cylindrical coils (Fig. 14) 
are: Om= 140 cm.; 6i=4.5 cm.; 6, =3 cm.; a = l cm. The maximum 
useful flux is 1.03 megalines. Show that the relationship between the 
height I of the winding and the percentage reactive voltage drop is 
xl = lQ6, Assume fc = 1.10. 

Prob. 2. Referring to the preceding problem, what is the permeance 
of the space between the outside low-tension coil and the high-tension 
coil, per centimeter of the height of the coils, and what is the effective 
value of the flux density in this space, at full load ? 

Ans. 197.5 perms per cm., 3420/Z lines per sq.cm. 

Prob. 3. Each leg of a core-type transformer is provided with six 
flat high-tension coils of 530 turns each, interposed with the same 
number of low-tension coils of 40 turns each, one of the low-tension 
coils being split in two and placed at the ends. The high-tension coils 
are wound of 3 mm. round wire, 53 layers, 10 turns per layer (6i = 3 cm.) ; 
the low-tension coils are wound of 8 mm. square wire, in 20 layers, 2 turns 
per layer (62 = 1.6 cm.). The distance between the coils is 20 mm. 
Taking the inductance of this transformer to be unity, calculate the 
relative inductances of the transformer when the high-tension winding 
is divided into three coils and also into two coils, assuming fc, I, and 
a to be the same in all cases. Ans. 2.55; 4.66. 

Prob. 4. Solve the preceding problem, taking into account the 
change in A:. Ans. 2.42; 4.14. 

Prob. 6. The following results were obtained from a short-circuit 
test on a 22/2-kv., 25(X>-kva., 60-cycle, shell-type transformer, with 
flat coils: With the high-tension winding short-circuited, and full rated 
current flowing through the low-tension winding the voltage across 
the secondary terminals was 73.5 v., and the wattmeter reading was 
27 kw. The transformer winding consists of 12 high-tension coils of 
100 timis each, and of 11 low-tension coils interposed between the 
high-tension coils, together with 2 half-coils at the ends. The dimen- 
sions of the coils are: Om =2.6 m.; 1 = IS cm.; 61 = 16 mm.; 62=* 10 mm.: 



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Chap. XII] INDUCTANCE OF WINDINGS * . 217 

a =12 mm. Calculate the correction factor A;' in formula (147). Hint: 
Eliminate the ir drop, using the wattmeter reading. Ans. About 1.05. 

Prob. 6. What is the greatest permissible thickness of the coils 
of a 60-cycle transformer, if the reactive drop must not be larger than 
three times the resistance drop? The ducts a are 1 cm. The space 
factor of the copper in each coil is 0.55. The primary coils and the 
whole coils of the secondary are of the same thickness. fc=0.98. 
Hint; Assume Omt I, and rii and show that they cancel out. 

Ans. 6=3.66 cm. 

Prob. 7. In order to provide a better cooling, and at the same 
time save on insulation, two flat high-tension coils are frequently placed 
side by side, with a small air-space in between, and in the same way 
the low-tension coils may be subdivided. Show that no leakage flux 
passes in the space between the two adjacent coils which belong to 
the same winding, so that the inductance of the winding is not appreciably 
increased by these spaces, and can be calculated as if these spaces did 
not exist.^ 

Prob. 8. Compare the formulaB given above and the numerical values 
of transformer leakage reactance obtained therefrom with the formulaB 
and data given in the Standard Handbook for Electrical Engineers, Do 
this for any transformer, the dimensions of which are available. 

Prob. 9. The equivalent reactance of a transformer is Xi reduced 
to the primary circuit, and is X2 reduced to the secondary circuit. All 
the primary coils are connected in series, and all the secondary coils 
are also in series. Show that these equivalent reactances become 
Xi/Ci^ and Xi/Ct^ respectively, when the primary winding is divided 
into Ci branches in parallel, and the secondary winding is divided into 
€2 branches in parallel. The division is supposed to be made in such 
a way as to keep the m.m.fs. in the adjacent coils balanced. Hint: 
If the equivalent reactance of one primary branch is Xi, that of Ci 
branches in series iaXi=XiCif and that of Ci branches in parallel is Xi/ci 
no matter whether the secondary coils are connected in series or in 
parallel. 

Prob. 10. Prove that the equivalent reactance of a transformer 
is the same, whether the coils are in series or in parallel, provided that 
the total number of turns in series is the same. 

Note: Sometimes, because of the difficulty in using heavy con- 
ductors, it is desirable to multiple the coils. In such case, the parallel 
coils miLst have the same nimiber of turns, and they should be sym- 
metrically arranged, so as to prevent exchange currents. 

66. The Equivalent Leakage Penneance of Armature Windings. 
In certain problems relating to the design and operation of 
electrical machinery it is necessary to calculate the inductance 

^ This fact has been verified experimentally; see Arnold, Wechselstrom' 
technik, Vol. 2 (1910), p. 29. 



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218 



THE MAGNETIC CIRCUIT 



[Art. 65 



of the annature windings. This inductance aflfects the per- 
formance of the machine, because the leakage fluxes created 
by the armature currents induce e.m.fs. in the machine. Such 
leakage fluxes are shown in Figs. 23 and 36, in an induction 
machine and a synchronous machine respectively. For purposes 
of theory and computation these leakage fluxes are usually 
subdivided into three parts, namely: 

(a) Leakage fluxes linked with the parts of the windings 
embedded in the armature iron (Figs. 36 and 54). These paths 
are closed partly through the slots, and partly through the tooth- 
tips (slot leakage and tooth-tip leakage). 

(6) Leakage fluxes linked with the parts of the armature 
windings in the air-ducts. 



-' N 



Cpp 

s 



s 



^N 



Cpp 

s 



*« le *• le 

Fig. 52. — ^Undivided end-connections. Fig. 53. — Divided end-connections. 

(c) Leakage fluxes linked with the end-connections of the 
armature windings (Figs. 52 and 53). 

Usually the fluxes (a) and (6) are merely distortional com- 
ponents of the main flux of the machine, and only the fluxes 
(c) have a real existence. 

It will be readily seen that the paths of the tooth-tip leakage 
and of that around the end-connections are too complicated 
to allow the corresponding permeances to be calculated theo- 
retically. For this reason, various empirical and semi-empirical 
formulae are used in practice for estimating the leakage inductance 
of annature windings, the coefiicients in these formulae being 
determined from tests on similar machines. 

The most rational procedure is to express the inductance 
through the equivalent permeance of the paths, as defined by 
eq. (106a) in Art. 58. Let there be Cpp conductors per pole 



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Chap. XIll INDUCTANCE OF WINDINGS 219 

per phase, such as are indicated for instance in Fig. 15. Then 
the inductance of a machine, per pole per phase, is given by the 
equation 

Lpp=(7pp2(Peg, (148) 

where (P^q is an empirical value of the equivalent permeance 
per pole per phase. This fonnula presupposes that all the 
partial linkages are replaced by the equivalent complete linkages 
embracing all the Cpp conductors. Moreover, the value of (Peg 
is such as to take into account the inductive action of the other 
phases upon the phase under consideration. The total inductance 
of the machine, per phase, depends upon the electrical connec- 
tions in the armature winding. If all the p poles are connected 
in series, the foregoing expression for Lpp must be multiplied 
by p; if there are two branches in parallel, the inductance 
of each branch is ipLpp, and the combined inductance of the 
whole machine is idpLpp) = ipLpp. 

The leakage inductance of a machine is usually determined 
by sending through it an alternating current of a known frequency, 
under conditions which depend upon the kind of the machine 
(the field to be removed in a synchronous machine, and the 
armature to be locked in an induction machine). From the 
readings of the current of the applied voltage and the watts 
input, the reactance x of the machine is calculated (after elim- 
inating the ohmic drop). Then, knowing the frequency / of the 
supply and the number of poles of the machine, the inductance 
Lpp=x/(27rfp) per pole is calculated. Substituting into eq. 
(148) this value of Lpp and the known number of conductors 
Cpp, the equivalent permeance (Peg per pole per phase is deter- 
mined. By performing such tests on machines built on the 
same punching, but of different embedded lengths, the permeance 
due to the embedded parts of the winding is separated from 
that due to the end-connections; the values so obtained are then 
used in new designs. 

The leakage permeances in the embedded parts are pro- 
portional to the widths of these parts in the direction parallel 
to the shaft, in other words, to the length of conductors which 
are surrounded by the leakage lines. Experiment shows that 
the permeance of the paths in the air-ducts and around the 
end-connections is also approximately proportional to the lengths 



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220 THE MAGNETIC CIRCUIT [Art. 65 

of the corresponding parts of the armature coils. Since all these 
permeances are in parallel, (Peq is equal to their sum, or 

^eg=^/i»+^a'^a + «^e^ .... (149) 

Here the letter I denotes the lengths in cm. of the coil, or, what 
is the same thing, the width of the paths of flux. The subscripts 
i, a, and e refer to the iron, the air-ducts, and the end-connections 
respectively. Thus, li is the semi-^aet length of the core, that is, 
the length exclusive of ducts but inclusive of the space between 
the laminations.^ The corresponding permeance per centimeter is 
CP/. The sign "prime " signifies that the corresponding (P's refer 
to one centimeter width of path. 

The coefficient a is equal to 1 when the end-connections 
are arranged as in Fig. 52, and a=i when they are arranged 
according to Fig. 53. Namely, in the first case the number of 
conductors Ce per group of the end-connections is equal to the 
number Cpp in the embedded part. In the second case Ce is 
equal to ^Cpp. In the first case there are as many groups of 
end-connections per phase as there are poles; in the second 
case there are twice as many groups per phase as there are poles, 
so that two groups (one pointing to the right and one to the left) 
must be counted per pole. Thus, with undivided end-con- 
nections, the inductance is Cp^G^eht while with divided end- 
connections it is (iCpp)2(Pe'.2Ze=Cpp2. J(Pe7e. This accounts for 
the value of a = i, in formula (149), and shows that the inductance 
due to the end-connections is reduced twice by subdividing 
them into two groups. When estimating the leakage reactance 
it is therefore necessary to know the exact arrangement of the 
end-connections. 

In preliminary calculations, before the armature coils are 
drawn to scale, the length l^ of the end-connections in a full- 
pitch winding is usually assumed to be equal to about l.Sr, where 
T is the pole pitch. For fractional-pitch windings, le varies 
roughly as the winding-pitch, or le= 1.5 l^r (see Art. 29). 

* The semi-net length is used in getting the leakage penneance in the 
slot, because the flux spreads as it comes out of the iron almost immediately. 
The spaces between the laminations do not afifect the density in the air 
because they aie so smaU as compared to the dimensions of the slot. 



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Chap. XII] INDUCTANCE OF WINDINGS 221 

Substituting the value of 6^eq from eq. (149) into eq. (148) 
we obtain 

^pp=C^pp'(^A+^a'^a+«^;U10-« henrys. . (150) 

In the following three articles formula (150) is applied to 
the calculation of the leakage reactance of 

(o) Induction machines; 

(6) Synchronous machines; 

(c) Coils undergoing commutation in a direct-current machine. 

In each case somewhat difiFerent values of the unit permeances 
(P' are used, because of the diversity of the magnetic paths. 

66. The Leakage Reactance in Induction Machines. It is 

explained in Art. 35 and shown in Fig. 23 that the actual flux 
in a loaded induction machine is the resultant of three fluxes, 
of which the useful flux is linked with both the primary and 
the secondary windings. The component fluxes 0^ and ^^y 
linked with the primary and secondary windings respectively, 
are called the leakage fluxes. They induce in the windings e.m.fs. 
in quadrature with the corresponding currents, and these e.m.fs. 
have to be balanced by a part of the terminal voltage. Con- 
sequently, that part of the applied voltage which is balanced by 
the useful flux is reduced; in other words, the useful flux and 
the useful torque are reduced with a given current input. As 
a matter of fact, the maximum torque and the overload capacity 
of an induction machine are essentially determined by its leakage 
fluxes, or what amounts to the same thing, its leakage inductances. 

Knowing the primary and secondary leakage reactances, the 
actual induction machine is replaced by an equivalent electric 
circuit (or a circle diagram is drawn for it), after which its 
performance can be predicted at any desired load. The problem 
here is to determine the values of these leakage reactances and 
inductances, from the given dimensions of a machine. The 
rest of the problem is treated in the Electric Circuit. 

The leakages fluxes, which are indicated schematically in 
Fig. 23, are shown more in detail in Fig. 54. The primary 
conductors in one of the phases and under one pole are marked 
with dots, and the adjacent secondary conductors are marked 
with crosses, to indicate the currents which are flowing in them. 

Assuming the rotor to be provided with a squirrel-cage 



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222 



THE. MAGNETI9 CIRCUIT 



[Art. 66 



winding, the distribution of the secondary currents is prac- 
tically an image of the primary currents. Neglecting the mag- 
netizing ampere-turns necessary for establishing the main or 
useful flux, the secondary ampere-turns per pole per phase are 
equal and opposite to the primary ampere-turns. A similar 
assumption is also made in the preceding article, in the case of 
the transformer. This assumption is not as accurate in the 
case of an induction machine, because here the magnetizing 
current is proportionately much larger, due to the air-gap; 
nevertheless, the assumption is sufficiently accurate for most 




Fig. 54'. — ^The slot and zig-zag leakage fluxes in an induction machine. 

practical purposes. Even if the magnetizing current is equal 
to say 25 per cent of the full-load current, the difference between 
the primary and the secondary ampere-turns should be less 
than 10 per cent, because the magnetizing current is considerably 
out of phase with the secondary current.^ 

With this assumption, the primary and the secondary current 
belts shown in Fig. 54 may be considered as two sides of a narrow 
fictitious coil which excites the leakage flux, causing it to pass 
circumferentially along the active layer .2 Neglecting the mutual 

' See the circle diagram of an induction motor, for instance, in the author's 
Experimental Electrical Engineering^ Vol. 2, p. 167. 

* Although the secondary frequency is different from the primary, with 
respect to the revolving rotor it is the same as the primary frequency with 
respect to the stator. Let s be the slip expressed as a fraction of the primaiy 
frequency. Then the speed of the rotor is (1— s), and the frequency of the 
secondary currents with respect to a fixed point on the stator is s + (1 — a) = 1. 



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Chap. XII] INDUCTANCE OF WINDINGS 223 

action of the consecutive phases, the length of this flux is approx- 
imately r/m, where t is the pole pitch and m is the number 
of the stator phases. Part of the flux is linked with the primary 
current belt, and part with the secondary belt. The conditions 
are essentially the same as between the transformer windings 
Pi and Si in Fig. 50. Knowing the equivalent permeances 
of the individual paths the inductance can be calculated from 
eq. (150). 

In an induction machine with a squirrel-cage rotor the total 
leakage in the embedded part may be resolved into three com- 
ponents shown in Fig. 54, namely : 

(1) The primary slot leakage, (^«i; 

(2) The secondary slot leakage, 0^2 f 

(3) The tooth-tip or zigzag leakage, 0g. 

The fluxes 0si and (^,2 are alternating fluxes of the frequency 
of the corresponding currents. The zigzag flux 0g varies according 
to a much more complicated law, because the permeance of its 
path changes from instant to instant in accordance with the 
relative position of the stator and rotor teeth; compare posi- 
tions (1) and (2) in Fig. 23. Moreover, Fig. 54 shows only the 
simplest case, which never occurs in practice, namely; when 
the stator tooth pitch is equal to that in the rotor. In reality, 
the two pitches are always selected so as to be different, in order 
to avoid the motor sticking at sub-synchronous speeds (due to the 
higher harmonics in the fluxes and in the currents). Therefore, 
the paths of the zigzag leakage flux are much more complicated 
than is shown in Fig. 54, and in calculations the average per- 
meance of the zigzag path is used. 

In a machine with a phase-wound rotor the main flux is 
further distorted, due to the fact that the primary and secondary 
phase-belts are not exactly in space opposition at all moments. 
While the total m.m.fs. of the primary and secondary are 
balanced, there is a local unbalancing which changes from 
instant to instant. This distortion is the same as if it were due 
to an additional leakage, which was named by Professor C. A. 
Adams the belt leakage. i This part of the leakage usually 
constitutes but a small part of the total leakage, and will not 
be considered here separately. Those interested are referred to 

* C. A. Adams, The Leakage Reactance of Induction Motors, Trans, 
Intern, Eledr. Congress, St. Louis, 1904, Vol. 1, p. 711. 



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224 THE MAGNETIC CIRCUIT [Art. 66 

the original paper and to the works mentioned at the end of 
this article. 

Let there be Cppi conductors per pole per phase in the stator 
winding; then the fictitious coil (Fig. 54) made up of the 
primary and secondary conductors has Cppi turns, when reduced 
to the primary circuit. This is because the secondary winding 
can be replaced by an equivalent winding with a " one to one " 
ratio, that is, with the same number of conductors as the 
primary winding. In this case, the secondary current is equal 
to the primary current (Art. 446). Therefore, eq. (150) can 
be made to give the equivalent inductance, including the pri- 
mary inductance and the secondary inductance reduced to the 
primary circuit, provided that the permeances of the paths 
linking with the secondary conductors are included in the values 
of (P"s. Such is naturally the case when the values are deter- 
mined from a test with the rotor locked.- 

Extended tests have shown that in a given line of machines 
the permeance (P/ is inversely proportional to the peripheral 
length of the equivalent leakage flux, that is, inversely proportional 
to (r/m), where t is the pole pitch and m is the number of 
primary phases.* This shows that the permeance per centi- 
meter of peripheral length of the active layer is fairly constant, 
in spite of different dimensions and proportions, as long as 
these are varied within reasonable limits. Thus 

(P/=(P/7(r/m), 

where (P/' is the leakage permeance of the active layer in the 
embedded part per one centimeter of axial length and per centi- 
meter of the peripheral length of the path. Thus, the final 
formula for the equivalent leakage inductance of an induction 
machine, per pole per phase, reduced to the primary circuit, is 

LeaPP==Cpp,^[(Prii/(T/m) + (P^'l^+a(P/le]10-^ henrys. (151) 

In this formula the following average values of unit per- 
meance may be used for machines of usual proportions, unless 
more accurate data are available. 

1 H. M. Hobart, Electric Motors (1910), table on p. 397. The values 
for (Pi" given below have been computed from this table, and the results 
multiplied by 2, because the table gives the values of the primary per- 
meances only. 



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Chap. XII] INDUCTANCE OF WINDINGS 225 

The equivalent permeance (P/' of the embedded part in perms 
per centimeter of the semi-net axial length of the machine, and 
per centimeter of peripheral length of the air gap, is for 

Open slots Half-open slots Completely closed slots 

11.5 14.5 18. 

The equivalent permeance around the end-connections, and 
around the parts of the conductors in the air-ducts, decreases 
with the increasing number of slots per pole per phase, for the 
same reason that the slot permeance decreased. In induction 
motors usually at least three slots are used per pole per phase, 
and under these conditions Mr. H. M. Hobart uses (P/ = 0,S 
perm per centimeter, with phase-wound rotors, and (P/ = 0,6 
perm per centimeter with squirrel-cage rotors. (P„' may be 
taken in all cases equal to 0.8 perm per cm. The lengths le 
and l^ are always understood to refer to the stator winding. 

The foregoing data refer to machines with full-pitch windings 
in the stator and in the rotor. With a fractional pitch winding 
the equivalent leakage permeances are somewhat smaller, due to 
longer phase belts and to the mutual induction of the over- 
lapping phases. Let the winding-pitch factor (Art. 29) of the 
primary winding be kj^i and that of the secondary winding 
k^2' Then the leakage inductance of the machine, calculated 
for a full-pitch winding (but for Ze=1.5I^T), is multiplied by 
*wi-*u;2« This is an empirical correction, which is accurate 
enough for ordinary practical purposes. In reality, of two 
machines designed for a given duty, one with a fractional pitch 
winding and the other not (but otherwise both alike) the first 
often has a higher inductance than the second. This is because 
more turns are required with the fractional pitch winding, if the 
flux densities in the iron and in the air-gap are to be the same 
in both cases.i 

With the data given above the calculation of the leakage 
reactance of a given induction motor is quite simple, and one 
engaged regularly in the commercial design of induction-motors 
can obtain sufficiently accurate data for their design or for the 

* For a theoretical and experimental investigation of the effect of a 
fractional pitch upon the leakage reactance in induction machines see 
C. A. Adams, Fractional-pitch Windings for Induction Motors, Trans. Amer. 
Inst, Elec. Engrs., Vol. 26 (1907), p. 1488. 



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226 THE MAGNETIC CIRCUIT [Abt. 66 

computation of their performance, provided that he adapts the 
numerical values of the unit permeances to each individual case, 
on the basis of his previous experience. Many authors have 
given theoretical formulae for the leakage inductance of induction 
machines.! These formulae, curves, and methods, while useful 
in accurate work, are too elaborate to be quoted here; at any rate 
they are of interest only to a specialist in design. Two examples 
of the theoretical calculation of leakage inductance are given 
below, in order to show the student the general method used, 
and thus introduce him into the literature on the subject. 

(a) A Theoretical CaLcuLation of the Slot Leakage Permeance. 
We shall calculate the equivalent permeance for the half closed 
slot (Fig. 55), an open slot being a special case of it. With 
the notation shown in sketch, and with Spp slots per pole per 
phase, we have: 

(Pc'=^fi[b2/s2+h/s3+b4'/s4]/Spp; d(Pp'=-/idx/(s4Spp), 

and nx=CppX/b4. Hence 






n^d(Pp' = ifiCpp^b4/84Spp, 

'0 

and 

^.pp'=C'pp2./£[62/S2 + 63A3+(J64 + V)A4]/5pp. . (152) 

The equivalent permeance of one slot, per unit of the semi- 
net axial length of the machine is 

(P/^filh/sz + h/ss+i^b^+b^^/s^]. . . . (153) 

This shows that the slot permeance depends only upon the 
proportions of the slot and not upon its absolute dimensions. 

(6) A Theoretical Calculation of the Zigzag Leakage Permeance. 
The calculation of the zigzag leakage permeance is simple only 

* C. A. Adams, loc, cit.; also Trans. Amer. Inst. Elec. Engrs., Vol. 24 
(1905), p. 338; ibid., Vol. 26 (1907), p. 1488. Hobart, Electric Motors, 
(1910), Chap, xxi; Arnold, Wechselstromtechnik, . Yo\. 5, part 1 (1909), pp. 
49-54; R. Goldschmidt, Appendix to his book on The AUemating Current 
Commutator Motor (1909); R. E. Hellmimd, Practische Berechnung des 
Streuungskoeffizienten in Induktionsmotoren, \Elektrotechnische Zeiischrift, 
Vol. 31 (1910), p. 1111 and 1140; W. Rogowski, Zur Streuung des Dreh- 
strommotors, ibid., pp. 356, 1292, and 1316. See also an extensive series 
of articles by J. Rezelman in La Lamihre Electrique, 1909-1911, and in 
the (London) Electrician. 



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Chap. XII] 



INDUCTANCE OF WINDINGS 



227 



when the stator tooth-pitch and the rotor tooth-pitch are equal 
to each other; otherwise the paths become too complicated for 
mathematical analysis. Since the position of the secondary teeth 
varies with respect to the primary teeth, the permeance of the 
zigzag leakage also varies, and it is necessary to take its average 
value over one-half of the tooth-pitch il, that is, between the posi- 
tions (1) and (2) in Fig. 23. In some intermediate position (Fig. 
55), determined by the overlap y, the reluctance of the path /, 
per unit of semi-net axial length of the iron, is a/i/iy) rels. 




Fig. 55. — The notation used in the calculation of the slot and zigzag leakage 

The reluctance of the path g is a//i(<2 ~^i "~2/) rels. The com- 
bined reluctance of / and g is equal to the sum of the foregoing 
expressions. The permeance of / and g in series, being the 
reciprocal of the combined reluctance, is 

The permeance in question varies according to this law, for the 
positions of the secondary tooth between t/=J(<2~^i) ^^^ 2/=0> 
the tooth moving to the left. From y = to y=i(t2—Si—X) the 
permeance is practically equal to zero, because the secondary tooth 
bridges over the primary slot no more. The student is advised 



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228 THE MAGNETIC CIRCUIT [Art. 66 

to mark the positions of the rotor teeth on a strip of paper and 
to place them in different positions with respect to the primary 
slot in order to see the variations in the overlap. Thus, the 
average value of the zigzag permeance per tooth pitch is 



"■-U 



,^ * " (Pydy=fi(t2Si)^/(QaX) perms/ cm. (154) 



Prob. 11. Draw a sketch similar to Fig. 54, but with the rotor 
tooth-pitch different from that in the stater, and indicate roughly the 
genersJ character of the paths of the zigzag leakage. 

Prob. 12. Show that increasing the number of slots per pole in a 
given machine does not alter materially the slot leakage, but reduces 
considerably the zigzag leakage. 

Prob. 13. Calculate the equivalent leakage inductance per phase 
of a three-phase, 10-pole, induction machine, with 15 slots per pole in 
the stator, and a phase-wound rotor. Both windings have 100 per 
cent pitch, the slots are semi-closed on both punchings, the individual 
stator coils in each phase are connected in series, and there are 20 con- 
ductors in each stator slot. The bore of the machine is 110 cm., the gross 
length of the core is 30 cm. ; there are three vents of 8 mm. each. The 
end-connections are arranged according to Fig. 53. Ans. 57.5 mh. 

Prob. 14. The design of the machine in the preceding problem 
has been modified in the following respects: The rotor is provided 
with a squirrel-cage winding, the winding pitch in the stator is shortened 
to 80 per cent, the stator slots are made open, and the end connections 
are divided, as in Fig. 54. What is the inductance of the machine? 

Ans. 43.3 mh. 
Prob. 16. Check the values of (P/' given in the text above with 
those in Hobart's table. 

Prob. 16. For the usual limits of proportions of slots, teeth, and 

air-gap calculate the values of (P/' 
-^ » 1^ from eqs. (153) and (154) and com- 

j-i-i . . pare the residts with the average 

^•-y^v. experimental values given in this 

/^ ^ text. 

I da \ Prob. 17. Calculate the equi- 

^ ^^^^i2MMa- valent leakage permeance of a round 

C xT ''''''''y X giQ^ (Pig 5gj^ assuming the con- 

[ ^^J — ^ I ductors to completely fill it, and 

"^ / the lines of force to be straight lines. 

^ ; •' Hint: Select the angle a as the inde- 

FiG. 56.— A semi-closed round slot, pendent variable, and integrate eq. 

(106) between a = and a = n. 
See Arnold, WechseUtromtechniky Vol. 4 (1904), p. 44. 

Ans. (Pa' = jt£(0.623 +6/s) perms per cm. 



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Chap. XU] INDUCTANCE OF WINDINGS 229 

67. The Leakage Reactance in Synchronous Machines. The 

physical nature of the armature reactance in a synchronous 
machine is explained in Art. 46; the influence of this reactance 
upon the performance of a machine is shown in Figs. 37, 38, 40, 
and 41. The problem here is to calculate the numerical value 
of this reactance for a given machine, using eq. (150) with 
empirical coefficients (P\^ 

It may also be stated here that for standard machines, partic- 
ularly in preliminary estimates, the ix drop is sometimes taken 
as a certain percentage of the rated voltage of the machine 
instead of estimating the inductance from formula (150). In 
synchronous generators the ix drop at the rated volt-ampere 
load varies from 5 to 10 per cent of the rated terminal voltage. 
In synchronous motors, where some inductance is useful, the 
ix drop ranges from 8 to 15 per cent of the rated voltage. For 
60-cycle machines, and for machines with a comparatively large 
number of armature ampere-turns, values must be taken nearer 
the higher limit. For 25-cycle machines, and for machines 
with a comparatively small number of armature ampere-turns, 
values must be taken nearer the lower limit. A considerable 
error in estimating the value of ix has but little effect upon the 
calculated performance at unity power factor, because the vector 
ix is then perpendicular to e (Figs. 37, 38, 40 and 41). However, a 
considerable error may be introduced at lower values of the 
power factor if the reactive drop ix has not been estimated 
with a sufficient accuracy. 

The values of (P' for synchronous machines are different 
from those given above for induction machines, because of the 
absence of any secondary current-belts. Parshall and Hobart2 
give the following values for 6^/ : 

* For a theoretical calculation of the coeflScient (P', see Arnold, Wechsd- 
stramtechnik, Vol. 4 (1904), pp. 41-52; Hawkins and Wallis, The Dynamo, 
Vol. 2 (1909), pp. 901-904. For a comparison between the calculated and 
actually measured values see an extended series of articles by J. Rezelman 
in La Lumih-e Electrique, 1909-1911, and in The (London) Electrician, 

^ Electric Machine Design (1906), p. 478. These values are corroborated 
by those obtained by Pichelmayer; see his Dynamohau, 1908, pp. 208 and 
504. Pichelmayer's values for (P^' (which he denotes by t) are somewhat 
high because the end-connection leakage is not considered separately. 
Arnold's values, given in his Wechselstromtechniky Vol. 4, p. 280, ^ould be 
used with discretion, because they apply to a different formula; namely, 



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230 THE MAGNETIC CIRCUIT , [Art. 67 

Uni-coil windings in open slots .... 3 to 6 perms per cm. 
Thoroughly distributed windings in 

open slots 1.5 to 3 " " 

Uni-coil windings in completely 

closed slots 7 to 14 " " 

Thoroughly distributed windings in 

completely closed slots 3 to 6 " " 

The much larger values of (p/ for closed slots, as compared 
to those with open slots, were to be expected because the bridge 
which closes the slot offers a path of high permeance. The 
lower values for windings distributed in several slots per pole 
per phase, as compared to uni-slot windings, are due to the 
fact that the partial linkages become more and more pronounced 
as the winding is distributed into a larger number of separate 
coils, and also because the length of the paths is greater. This 
is somewhat analogous to splitting a transmission line into two 
or more lines; see prob. 21 in Art. 61. The greatest reduction 
in the value of the inductance results when the number of slots 
is increased from one to two; a further subdivision is of much 
less importance. For instance, if the permeance (?( with a 
uni-slot coil is 7, then dividing the same coil into two slots 
reduces the permeance to less than 5. On the other hand, a 
change from four to five slots per phase per pole would hardly 
reduce the equivalent permeance more than from say 3.5 to 3.4. 
The data in the table above give rather a wide range from which 
to select a value of (Pi for a particular machine, and the designer 
must exercise his judgment as to whether his machine will have 
a permeance nearer the upper or lower limit. This judgment 
comes with experience, by comparing the predicted performance 
of machines with that actually observed. 

The values of (PJ^ and (P^ depend upon the number of coils 
per group, in other words, upon the number of slots per pole 
per phase. Until more accurate and detailed data are available^ 

he considers separately the equivalent permeance (P^ of eojch &\ot, instead 
of the group of slots per pole per phase. Thus, his formula for the leakage 
inductance per pole, with our notation, is Lpp^SppC^'^CP^^ where Spp is 
the number of slots per pole per phase, and C, is the number of armature 
conductors per slot. The values of unit permeance, which he gives and denotes 
by A, refer to this formula. 



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Chap. XII] INDUCTANCE OF WINDINGS 231 

we shall assume (Pj = (Pay and use the following values, based 
upon Mr. Hobart's experiments: ^ 

Number of slots per pole per phase .1 2 3 more than 3 
(Pj = (Pfj^y in perms per centimeter ..0.8 0.7 0.6 0.5 

With a fractional-pitch winding the inductance is some- 
what reduced (see the end of the preceding article). As an 
empirical correction, the inductance calculated for a full-pitch 
winding may be multiplied by the winding pitch factor kw* 

The leakage reactance of the armature cannot be calculated 
from a short-circuit test, because the short-circiut current is 
essentially determined by the direct armature reaction. A 
difference of 50 or even 100 per cent in the armature reactance 
would change the short-circuit current by only a few per cent. 
A much closer approximation is obtained from the so-called 
air-characteristic,^ Namely, it has been found by numerous experi- 
ments that the armature inductance, when the field is revol- 
ving synchronously, is nearly equal to the armature inductance 
with the field completely removed, and the armature supplied 
with alternating currents from an external source. The air- 
characteristic is the relation between the current and the voltage 
under these conditions. Eliminating the ohmic drop, the induct- 
ance of the machine is easily calculated ^ and the value so found 
can be used in the prediction of the performance of the machine, 
Such an air-characteristic is easily taken in the shop or in the 
power house before the machine is completely assembled. From 
the three observed curves, namely, the no-load saturation c\irve, 
the short-circuit curve, and the air-characteristic, the perform- 
ance of a synchronous machine at any load can be predicted 
to a considerable degree of accuracy. 

Prob. 18. What is the inductance per phase of a 6-pole, 3-phase, 
turbo-alternator, the armature of which has the following dimensions: 
Bore, 1.2 m., gross length of core, 1.2 m., 20 air-ducts of 1 cm. each, 
90 open slots? There are 8 conductors per slot, the winding is of the two- 
layer t3rpe, the winding pitch is 11/15. Assume (Pi'= 2.5 perms /cm. 

Ans. 24.3 mh. 

Prob. 19. The inductance of the machine specified in the preceding 
problem was determined experimentally (from an air-characteristic), 

^ Joum, Inst. Electr. Eng. (British), Vol. 31, pp. 192 ff. 
' Pichelmayer, Dynamohau (1908), p. 207. 



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232 THE MAGNETIC CIRCUIT [Art. 68 

and compared to that measured on a similar machine, the gross length 
of the core of which was 80 cm. and which was provided with 12 ducts 
of 1 cm. each. The equivalent leakage permeance of the shorter machine 
was found to be 30 per cent less than that of the other machine. What 
are the actual values of (P/ and (Pe'(=(Pa') for both machines? 

Ans. (P/ =2.46, (P/ =(Pa' = 0.56 perm/cm. 

Prob. 20. An alternator has 3 slots per phase per pole, and the 
equivalent permeance is(Pt' = 1.8; 6*/ =0,6, What would be the value 
of the same constants per slot? Ans. 5.4 and 1.8. 

Prob. 21. A 3-phase alternator has 4 slots per phase per pole. 
If the coils were connected up for a 2-phase machine without change 
what would be the ratio of the new L to the old? Ans. 3 : 2. 

68. The Reactance Voltage of Coils undergoing Commuta- 
tion. Let Fig. 57 represent a part of the armature winding 
and commutator of a direct-current machine, with two adjacent 
sets of brushes. During the interval of time when an arma- 
ture coil, such as CD, is short-circuited by a set of brushes, 
the current in the coil is reversed from its full value in one 
direction to an equal value in the opposite direction. The coil 
is then said to undergo commutation. 

Under unfavorable conditions this reversal of current is 
accompanied by sparking between one of the edges of the brushes 
and the commutator. Unless a machine is provided with inter- 
poles, its output is usually limited by this sparking at the com- 
mutator. It is of importance, therefore, to have a practical 
criterion for judging the quality of commutation to be expected 
of a given machine. Numerous formulae and methods have 
been proposed for the purpose; all rational formulsB contain, 
as a factor, the inductance of the coils undergoing commutation, 
because this inductance determines essentially the law according 
to which the current is reversed with the time. For this reason, 
the subject of commutation is treated in this chapter, under the 
general topic of the inductance of windings. The method of 
calculation of the inductance and the criterion of commutation 
given below are due to Mr. H. M. Hobart.^ 

A description of the phenomenon of commutation. The phenom- 
enon of commutation may be briefly described as follows: 
Let, for the sake of explanation, the armature and the commu- 

^ See Hobart, Elementary Principles of Contimums-CurrerU Dynamo Design 
(1906), Chap. 4; also Parshall and Hobart, Electric Machine Design (1906), 
pp. 171-194. 



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Chap. XII] 



INDUCTANCE OF WINDINGS 



233 



tator be assumed to be stationary, and the brushes revolving in 
the direction of the horizontal arrow, shown in Fig. 57. Let 
the machine be provided with a multiple winding (lap winding), 
so that there are as many armature circuits and sets of brushes 
as there are poles. The current through each armature branch 



is, therefore 



/i=/M 



(155) 



where / is the total armature current and p the number of poles. 




Ckunmutator- 



WW ^WP 



>ositlve 
brush 



N^rative 
brush 



Direction of motion 
of the brushes 



Fig. 67. — ^Part of the armature winding, commutator, and brushes in a 
direct current machine. 



At each set of brushes two branches of the armature winding 
are connected in parallel, so that the current through each set 
of brushes is equal to 2/i With reference to Fig. 57, it will 
be seen that the two armature branches, X and F, which begin 
at each set of brushes, may be called, with respect to this set, 
the left-hand branch and the right-hand branch. 



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234 THE MAGNETIC CIRCUIT [Art. 68 

In the position of the positive brushes just preceding that 
marked 1, the coil CD is not short-circuited, and carries the full 
current I\, being the first coil of the right-hand branch. The 
lead d is idle, and the total current 2/i is delivered to the brushes 
through the leads c, m, and n. In the position of the positive 
brushes just after that marked 2, the coil CD is again not short- 
circuited, but is carrying the full current /i in the opposite direc- 
tion, being the first coil of the left-hand branch. 

In the positions of the brushes between 1 and 2 the coil CD 
is short-circuited by the brushes through the leads c and d, 
and the current in the coil changes graduall}' from +/i to — /i. 
If the coil possessed no inductance, the variation in the current 
would be practically determined by the contact resistance 
between the brush and the commutator, the resistance of the 
coil itself and of the leads being negligible (with carbon brushes). 
Under these conditions the current in the short-circuited coil 
would vary with the time according to the straight-line law, 
and the current density under the heels and the toes of the 
brushes would be the same. This is called the " pure resistance " 
commutation, or the perfect commutaton, because it is not accom- 
panied by sparking. Such a commutation is approached in 
machines with interpoles, when the effect of the inductance is 
correctly compensated for by the commutating flux (Art. 54). 

In reality, the short-circuited coU possesses a considerable 
inductance, which has the effect of electromagnetic inertia, 
retarding the reversal of the current. Consequently, at the 
beginning of the commutation period the lead d and the corre- 
sponding commutator segment do not carry their proper share 
of the current, which they would carry with a perfect com- 
mutation. At the end of the commutation period the current 
must then be reversed quickly, because the whole current must 
be transferred from the lead c to the other leads. If the inductance 
is considerable, the current in the lead c is still of a considerable 
magnitude when the toe of the brush is about to leave the 
corresponding commutator segment. Therefore, the last period 
of the reversal is accomplished through the air between the 
brush and the segment, in the form of an electric arc. This 
is known as the sparking at the brushes. Besides, during the 
last moments of reversal, the current density under the toe 
is much higher than the average density under the brush, and 



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Chap. XII] INDUCTANCE OF WINDINGS 235 

this high density causes a glowing at the edge of the brushes, 
making the commutation still less satisfactory. 

The average reactance e,m.f. in the coils of a fuU-pitch lap 
winding. For an empirical criterion of the quality of commutation 
Mr. Hobart takes t?ie average reactance voltage induced in the coil. 
This is a reasonable criterion, because the ratio of the maximum 
voltage occurring when the brush leaves a segment to the average 
voltage, wiU be more or less the same in machines of usual 
design constants. Of course, the average reactance voltage 
is only a relative criterion, to be used with great discretion, 
and applied only for comparison with machines which proved 
in actual operation to commutate satisfactorily. 

Let the inductance of an armature coil between two' adjacent 
commutator segments be Leq. The subscript eq (meaning equiva- 
lent) is added to indicate that the value of L includes not only the 
true inductance of the coil itself, but also the average inductive 
action of the coils which are undergoing commutation simultane- 
ously with it. Let the frequency of the current in the coil under- 
going commutation be / cycles per second. Then the current is 
reversed in a time i/. The flux changes during this time 
from +LeqI\ to —Leqii. Hence, according to the fundamental 
eq. (26) Art. 24, the average reactance voltage, which is taken 
as the criterion of commutation, is 

^,,e = 4/Leg/i (156) 

Li order to obtain a satisfactory commutation, the voltage 
Cave niust uot exceed a certain value, determined from actual 
experience with machines in regular operation. Mr. Hobart 
recommends values for Cave not to exceed 3 to 4 volts, provided 
that one does not depend upon the fringe flux of the main poles 
to facilitate commutation. 

The inductance Leq, which enters in the foregoing formula, 
is calculated according to the general formula (150), as follows: 
Assume first that there is no common flux or mutual induction 
between the coil under consideration and the other coils which 
are simultaneously short-circuited. Then, if g is the number 
of turns per commutator segment (in Fig. 57 q=l) we must put 
Cpp=q. This will give the inductance of one side of the coil, 
say C. To obtain the inductance of both sides, C and D, the 
result must be multiplied by 2, or Leq=2Lpp. 



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236 THE MAGNETIC CIRCUIT [Art. 68 

In reality there is a common flux which links with the coil 
under consideration and with the other coils undergoing com- 
mutation at the same time. This flux is changing with the 
time, and consequently it induces additional voltages in the 
coil CD. The induced e.m.f. depends upon the relative 
position of CD and the other coils (whether in the same slot 
or in the adjacent slots) and upon the rate of change of the cur- 
rent in each coil of the group. It would be too complicated for 
practical purposes to take all these factors into account with 
any degree of accuracy. Therefore, Hobart makes a further 
assumption, namely, that the current in all the coils, which are 
short-circuited at the same time, varies at the same rate and that 
the whole leakage flux is linked with all the coils of the group (Fig. 
57). 

Let s be the average number of coils simultaneously short- 
circuited under a set of brushes (the actual number varies from 
instant to instant). Consider a group of mutually influencing 
conductors, such as are shown at C or at D. One-half of the 
conductors in the same group are short-circuited by the positive 
brushes, the other half by the adjacent negative brushes. The 
total number of coils in each group is 2s, and since by assumption 
the current in all of them varies at the same rate, and all of 
the flux is linked with all of the coils, the equivalent inductance 
of the coil AB is 2s times larger than if this coil were alone. 
Thus for a multiple-wound armature 

Lea=2Lpp.2s=4sq^((Pi%+(Pa%+i(Pe%)X10-^henry8. (157) 

On the basis of Mr. Hobart's tests and until more accurate data 
are available, the following average values of the imit permeances 
may be used: tP/ = 4 and (Pa=(P/=0,S perms per centimeter. 

The frequency / which enters into formula (156) is calculated 
as follows: The time between the positions 1 and 2 of the brushes 
corresponds to one-half of one cycle, because during this interval 
the current changes from +/i to — /i. Let v be the peripheral 
velocity of the commutator, in meters per sec, let b be the thick- 
ness of the brushes, and 6' the thickness of the mica insulation 
between the commutator segments, both in millimeters. The 
time between the positions 1 and 2 of the brushes is (b—b') /lOOOv 
seconds. Hence 

/=5OO2;/(6-&0 cy./sec (158) 



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Chap. XIIl INDUCTANCE OF WINDINGS 237 

The number of simultaneously short-circuited coils varies 
periodically with the position of the brushes. Thus, in Fig. 57 
sometimes two and sometimes three coils are short-circuited 
by one set of brushes. On the average 

5=(6-60/a, (159) 

where a is the width of one commutator segment including the 
mica insulation. Thus, all the values which enter into the 
formula (156) are determined, and the reactance voltage for a 
given machine can be easily calculated. 

Formula (156) is used not only as a criterion of the commuta- 
tion, but also for the calculation of the flux density, under the 
interpoles where such are required. Namely, this flux density must 
be such that the average voltage induced by the commutating 
flux is approximately equal and opposite to the average reactance 
voltage; see Art. 24. prob. 6. A still closer compensation for the 
influence of the inductance is achieved by properly grading the 
commutating flux, so as to compensate not only for the average 
reactance voltage, but also to some extent for the instantaneous 
induced e.m.fs. 

The average reactance e,m.f, induced in some other direct cur- 
rent windings. With two-circuit wave windings two cases must 
be considered, namely, (a) when the machine is provided with 
only two sets of brushes, (6) when there are more than two sets 
of brushes. In the first case eq. (156) is used, where 

/i = i/, (160) 

and Leg is understood to comprise the short-circuited conductors 
under all the poles, per commutator segment. Let there be 
again q turns per coil, that is, per unit of winding per pair of 
poles. Since the corresponding conductors under all the poles 
are in series, we have that Leq=pLpp, The influence of the 
other simultaneously short-circuited coils is expressed as before 
by the factor 2s, where s is given by formula (159). Thus, for 
a two-circuit winding with two sets of brushes 

Le^=2psq2((Pi%+ (Pa'la+i(Pe%) X lO"! hcnrys. . (161) 

The frequency /is given as before by eq. (158). 

When more than two sets of brushes are used, the sets of 
equal polarity are connected in parallel by the stud connections 



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238 THE MAGNETIC CIRCUIT [Abt. 68 

outside the armature windings, and beside there are two short- 
circuiting paths through the coils undergoing commutation: a 
long path and a short path. Thus, the problem becomes 
indefinite, because it is not possible to tell the relative amoimts of 
the current through these different paths. Disregarding the short 
path, the criterion becomes the same as in the case of a machine 
with two sets of brushes only.^ This is on the safe side, and 
the commutation may be expected to be better than that cal- 
culated, or at the worst, as good. 

With multiplex windings, the expression for e^M is the 
same as that given above, provided that proper values are selected 
for /i, Sj and /. With regard to the latter quantity it must 
be remembered that 6' is much larger than the actual thickness 
of mica. Namely, with respect to the component winding under 
consideration the metal of the commutator segments belonging 
to the other component • windings is equivalent to insulation. 
This fact must not be lost sight of in choosing the correct value 
for V to be used in the expression (158). 

With fractional-pitch windings the reactance voltage is smaller 
than with the corresponding full-pitch winding, because the 
conductors short-circuited under the adjacent sets of brushes 
(Fig. 57) are situated in part or totally in different slots, and 
have a smaller common magnetic flux, or none at all. When 
the winding pitch is reduced considerably, s instead of 2s must 
be used in the preceding formulae ; otherwise a value between 
s and 2s must be chosen, according to one's judgment.^ 

Prob. 22. The armature of a 6-pole, 600-r.p.m., multiple-wound, 
direct-current machine has the following dimensions: Diameter, 85 cm.; 
gross length, 22 cm.; three air-ducts, 1 cm. each; 1008 face conductors. 

^ For an analysis of this case see C. A. Adams, Reactance E.M.F. and 
the Design of Commutating Machines, Electrical World and Engineeff Vol. 
46 (1905), p. 346. 

^ For an advanced and more scientific theory of commutation, see Arnold, 
Die Gleichstromaschine^ Vol. 1 (1906), pp. 354 to 513; in particular the 
approximate formula (170) on bottom of p. 498; also Vol. 2 (1907), chapter 
14. A simpler and more concise treatment will be found in Tomalen's 
Electrical Engineering. A good practical treatment wiU also be found in 
Pichelmayer's Dynamohau, pp. 86-118; it is considerably simplified as 
compared to Arnold's treatment, and is accurate enough for practical pur- 
poses, because the numerical values of unit permeances are known only 
approximately. 



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Chap. XII] INDUCTANCE OF WINDINGS 239 

The commutator diameter is 52 cm.; the number of segments, 252; 
the mica insulation is 1 mm. thick; the brushes are 15 mm. thick. 
What is the average reactance voltage when the total armature current 
is 320 amp.? Ans. 4.52 volts. 

Prob. 23. Show that the answer to the preceding problem would 
be nine times larger if by mistake the winding were assumed to be of 
the two-circuit type. 

Prob. 24. The peripheral velocity of a commutator is 18 meters 
per sec, the width of each segment (without mica) is 4.5 mm.; the 
thickness of the mica is 0.9 mm. The conunutator is to be used in con- 
nection with a duplex winding. What is the smallest permissible 
thickness of the brushes if the frequency of commutation must not 
exceed 800 cycles per sec.? Ans. 17.5 mm. 

Prob. 26. For a perfect commutation and for an imperfect one 
draw the following curves to time as abscissae: (a) the current in the 
short-circuited coO; (b) the currents in the leads c and d] (c) the 
current densities under the heel and toe of the brush. Take the width 
of the brushes to be equal to that of one commutator segment, and 
assume the mica insulation to be of a negligible thickness. 

Prob. 26. Show that the width of the. brushes has comparatively 
little net effect upon the commutation of a machine. 

Prob. 27. What flux density is needed in the interpolar zone in 
prob. 22 to secure perfect commutation? Ans. 2,23 kl./sq. cm. 



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CHAPTER XIII 

THE MECHANICAL FORCE AND TORQUE DUE TO 
ELECTROMAGNETIC ENERGY. 

69. The Density of Energy in a Magnetic Field. The reader 
is already familiar with the fact that a certain amount of energy 
is required to establish the flux within a magnetic circuit, and 
that this energy remains stored in the field. This stored energy 
may be conveniently thought of as the kinetic energy of vor- 
tices around the lines of force (Art. 3). Various expressions 
for the total stored electro-magnetic energy are given in Arts. 
57 and 58; the problem here is to find a relation between the 
distribution of the flux density and that of the energy in the 
field. 

Consider first the simplest magnetic circuit (Fig. 1) con- 
sisting of a non-magnetic material. According to the last eq. 
(99), the total energy stored in such a circuit is 

Tr=i*2Z/(^) joules, (162) 

if tf is in webers, I and A in cm., and /£= 1.257X10""^ henrys 
per cm. cube. The volume of the field is V^IA cubic cm. 
Since the flux density is uniform, the energy is also uniformly 
distributed, and the density of the energy is 

Denoting the density of the energy W/V by TT', and introducing 
the flux density B=0/Aj we get 

TF'=J52//£ joules per cu.cm. . . . (163) 

Either B or /jl can be eliminated from this expression by means 
of the relation B=ixH, so that we have two other expressions 
for the density of the energy: 

TF'=i/J?2; (164) 

W'=^BH (165) 

240 



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Chap. XIII] TORQUE AND TRACTIVE EFFORT 241 

Two more expressions for the density of the energy can be written, 
using the reluctivity v instead of the permeability pt. 

In a uniform field the preceding expressions represent the 
actual amounts of energy stored per cubic centimeter. In a 
non-uniform field TT' is the density of energy at a point, or the 
limit of the expression AW/ AY, This is analogous to what we 
have in the case of a non-uniform distribution of matter, where 
the density of matter at a point is the limit of the ratio of the 
mass to the volume. Thus, the total energy stored in a non- 
uniform field is 

W-^\vCwdY, (166) 

where the integration is to be extended over the volume of the 
whole magnetic circuit. Similarly, from eqs. (164) and (165) 
we get 

Tr=i;iJ^Vd7, (167) 

W^\^^UBdY (168) 

These expressions are consistent with eqs. (102) and (102a) 
as is shown in prob. 6 below. 

When li. is variable, the preceding formulae do not hold true, 
and the density of energy is represented by eq. (19), Art. 16. 

Prob. 1. Deduce an expression for the magnetic energy stored in 
the insulation of a concentric cable (Fig. 46), between the radii a and 6, 
the length of the cable being I cm. and the current t. Hint: For an 
infinitesimal shell of a radius x and thicknesss dx we have : H = i/2nx, 
and dV^27:zl dx, Ans. IF =0.23^1* log (6/a)10-« joules. 

Prob. 2. Check the answer to the preceding problem by means 
of eqs. (104) and (109). 

Prob. 3. In a concentric cable (Fig. 46) a«7 mm. and 6 is 20 mm. 
What is the density of the energy at the inner and outer conductors, 
when i« 120 amp.? 

Ans. 4.68 and 0.57 microjoules per cu.cm. 

Prob. 4. Deduce expression (110) from eq. (167). 

Prob. 6. Taking the data from the various problems given in this 
book as typical, show that ordinarily in generators and motors a large 
proportion of the total energy of the field is stored in the air-gap. 

Prob. 6. Show that eqs. (166) to (168) are consistent with eqs. 
(102) to (103tt). Solution : Take an infinitesimal tube of partial linkages 



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242 THE MAGNETIC CIRCUIT [Art. 70 

(Fig. 45). The energy contained in this tube is dW^\Mpd0p] but 
Mp= I Hdl, and d0=BdA. Since d(P is the same through all cross- 
sections of the tube, d<P can be introduced under the integral sign, and 
we have dW^i I HdlBdA=i I HBdV, the integration being extended 

over the volume of the tube. The total energy of the circuit is foimd 
by extending the integration over the volume of all the tubes of the 
field. The other equations are proved in a similar manner. 

70. The Longitudinal Tension and the Lateral Compression 
in a Magnetic Field. The existence of mechanical forces in a 
magnetic field is well known to the student. He needs only 
to be reminded of the supporting force of an electromagnet, of 
the attraction and repulsion between parallel conductors carrying 
electric currents, of the torque of an electric motor, etc. These 
mechanical forces must necessarily exist, if the magnetic field 
is the seat of stored energy. This is because, if we deform the 
circuit, we must in general change the stored energy and hence 
do mechanical work. The lines of force tend to shorten them- 
«selves and to spread laterally, so as to make the permeance of 
the field a maximum, with the complete linkages. Where there 
are partial linkages, it is the total stored energy that tends toward 
a maximum (Art. 57). This fact is entirely consistent with 
the hypothesis of whirling tubes of force, because the centrifugal 
force of rotation produces exactly the same effect, that is, a lateral 
spreading and a tension along the axis of rotation. A good 
analogy is afforded by a short piece of rubber tube filled with 
water and rotated about its longitudinal axis. 

(a) The Longitudinal Tension, Consider again the simple 
magnetic circuit (Fig. 1), and let it be allowed to shrink, due 
to the longitudinal tension of the lines of force, so as to reduce 
its average length by M, without changing the cross-section A. 
Let at the same time the current be slightly decreased so as to 
keep the same total flux as before. Let F/ be the mechanical 
tension along the lines of force, per square cm. of cross-section 
A; then the mechanical work done against the external fowes 
which hold the winding stretched is (F/ .A) JL The density 
of energy W^ remains the same because B is the same, but the 
total stored energy is decreased by TF'(AJZ), because the volume 
of the field is decreased by AM. Since the change was made 



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Chap. XIII] TORQUE AND TRACTIVE EFFORT 



243 



in such a way as to keep the total flux constant, no e.m.f. was 
induced in the winding during the deformation, and consequently 
there was no interchange. of energy between the electric and 
the magnetic circuit. Thus, the decrease in the stored energy 




Coll 



Fig. 68. — A lifting electromagnet. 

is due entirely to the mechanical work performed. Equating 
the two preceding expressions^ we have that 

F/^W'=iB2/fi=ifiH^ (169) 

If "FT' is in joules per cu.cm., F'< is in joulecens per sq.cm. (see 
Appendix I), so that in a rational system of units the mechanical 
stress per unit area is numerically equal to the density of the stored 



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244 THE MAGNETIC CIRCUIT [Art. 70 

energy. The physical dimensions of F' and W are also the 
same. 

If Ft is in kg. per sq.cm., jB in kilolines per sq.cm., and 
H in kiloampere-tums per cm., the preceding formula becomes, 
when applied to air, 

F/=52/24.7=ffVl5.6 (170) 

These formulae apply directly to the lifting magnet (Fig. 
58), and give the carrying weight per unit area of the contact 
between the core and its armature. The total weight which 
the magnet is able to support is 

Ft=AB^/2^.7=AHyi5.6kg., . . . (171) 

where A is the sum of the areas denoted by Si and /S2. Of course, 
H is taken for the air-gap, which is the only part of the circuit 
that is changing its dimensions when the armature is moved. 

(b) The Lateral Compression. Let now the simple magnetic 
circuit, be allowed to expand laterally by a small length Js in 
directions perpendicular to the surface of the toroid. Let F/ 
be the pressure (compression) exerted by the lines of force upon 
the winding, per sq. cm. of the surface of the toroid. Then 
the mechanical work done by the magnetic forces in expanding 
the ring against the external forces which hold the winding, 
is SFcJS, where S is the surface of the toroid. Let again the 
current be slightly decreased during the deformation, so as to 
keep the flux constant. No voltage is induced in the winding, 
and hence there is no interchange of energy between the electric 
and the magnetic circuit. Thus we can find F/, as we found 
the stress in the case of the tension, by equating the work done 
to the decrease in the stored energy. The stored energy is 
expressed by eq. (162), in which A is the only variable; hence 
by differentiating W with respect to A we get : 

JW^ -i0HJA/(fiA^)^ -i52(ZJA)//i. 

This is a negative quantity, because the stored energy decreases. 
But IJA represents the increase in the volume of the ring, so 
that lAA = SASj and consequently 

FJ^iBypL^iliH^^W'^F/ (172) 

In other words, the lateral compression is nummerically equal to 



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Chap. XIII] TORQUE AND TRACTIVE EFFORT 



245 



t?ie longitudinal tension, and both are numerically equal to the 
density of the stored energy. 

As an application of the lateral action, consider a constant- 
current or floating-coil transformer (Fig. 59), used in series 
arc-lighting. The leakage flux is similar in its character to that 
shown in Figs. 50 and 51. The lateral pressure of the leakage 
lines between the coils tends to separate them, acting against 
the weight of the floating coil. A part of this weight has to be 
balanced by a counter-weight Q because the electro-magnetic 
forces under normal operation are comparatively small. 

Since the currents are alternating, the force is pulsating, 




Primal^ Gbll* 



Primaiy OoU* 



Fig. 59. — ^A floating-coil constant-current transformer. 



itant-cu 



Potential A.C. 
Supply 



but is always in the same direction, tending to separate the coils. 
The average force depends upon the average value of H^^ in 
other words, upon the effective value of the current. According 
to eqs. (170) and (172), we have 

(Fc)ave=-(F,\^.S^Hef^S/15.&kg,, . . (173) 

where S is the area of the floating coil in contact with the flux. 
With the assumed paths for the lines of force, and neglecting 
the reluctance of the iron core, we have that 



Heff^nleff/lOOOl kiloamp.-tums/cm., 



(174) 



where I is the length of the lines of force in the air, in cm., and 
nieff is the m.m.f. of either coil. The force of repulsion is pro- 
portional to the square of the current, and is independent of the 



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246 THE MAGNETIC CIRCUIT [Art. 70 

distance h between the coils. Hence, a constant weight Q 
regulates for a constant current. When the coils are further 
from each other, the induced secondary voltage is less, on account 
of a much higher leakage flux. When the current increases 
momentarily, due to a decreasing line resistance, the coil is 
overbalanced and rises till the induced voltage and current fall 
to the proper value. Thus, the coil always floats at the proper 
height to induce the voltage needed on the line. 

Formulae (173) and (174) apply also to the mechanical forces 
between the primary and the secondary coils of a constant- 
potential transformer (Figs. 13 and 51). Under normal con- 
ditions these forces are negligible, but in a violent short-circuit 
the end-coils are sometimes bent away and damaged, unless 
they are properly secured to the rest of the winding. Such 
short-circuits are particularly detrimental in large transformers, 
having a close regulation, that is, having a very small internal 
impedance drop, and which are connected to systems of practically 
unlimited power and constant potential. As Dr. Steinmetz puts 
it, the closest approach to the appearance of such a transformer 
after a short-circuit is the way two express trains must look 
after a head-on collision at high speed. 

Another interesting example of the effect of the mechanical 
forces produced by a magnetic field is the so-called pinch phe- 
nomenon.^ The lines of force which surround a cylindrical con- 
ductor may be compared to rubber bands, which tend to com- 
press it. With a liquid conductor and large currents, such 
for instance as are carried by a molten metal in some electro- 
metallurgical processes, the pressure of the magnetic field is 
sufiicient to modify and to reduce the cross-section of the liquid 
conductor. This was first observed by Mr. Carl Hering and 
called by him the pinch phenomenon. In passing a relatively 
large alternating current through a non-electrolytic liquid con- 
ductor contained in a trough, he foimd that the liquid contracted 
in cross-section and flowed up-hill lengthwise in the trough, 
climbing up on the electrodes. With a further increase of 

* E. F. Northrup, Some Newly Observed Manifestations of Forces in the 
Interior of an Electric Conductor, Physical Review j Vol. 24 (1907), p. 
474. This article contains some cleverly devised experiments illustrating 
the pinch phenomenon, and also a mathematical theory of the forces which 
come into play. 



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Chap. XIII] TORQUE AND TRACTIVE EFFORT 



247 



current, this contraction of cross-section became so great at 
one point that a deep depression was formed in the liquid, with 
steeply inclined sides, like the letter V. 

In most cases of mechanical forces in a magnetic field, these 
forces and the resulting movements are due to the combined 



Slop 




ilt^ 



I i 



V 



•') 



Coil 



Cons 



Plunger 




Fig. 60. — ^A tractive electromagnet. 




Fig. 61. — ^Two bus-bars and their 
support. 



action of longitudinal tensions and transverse compressions and 
not to one of these actions alone. For instance, a loop of flexible 
wire, through which a large current is flowing, tends to stretch 
itself so as to assume a maximum opening, that is, a maximum 
permeance of the magnetic field linked with it. This action 
is due to both the longitudinal tension and the lateral pressure. 
In such cases the mechanical forces are best computed by the 
principle of virtual displacements explained in the next article. 



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248 



THE MAGNETIC CIRCUIT 



[Art. 70 



Prob. 7. Show that the required flux density in the air-gap of a 
lifting ele ctroma gnet (Fig. 58) can be calculated from the expression 
B = 15.7\/aF/Af in kl/sq.cm., where F is the rated supporting force, 
in metric tons, A is the area of contact in sq. dm., and a is the factor 
of safety. 

Prob. 8. Show that in an armored tractive magnet (Fig. 60) the 
tractive effort F varies with the air-gap « according to the law Fs' = 3.08 
kg-cm.. when the excitation is 2(X)0 amp .-turns and the cross-section 
of the plunger and of the stop is 12 sq.cm. Assume the leakage and the 
reluctance of the steel parts to be negligible. 

Prob. 9. Referring to the preceding problem, what is the true 
average pull between the values « = 1 and «-=4 cm., and what is the 
arithmetical mean pull? Ans. 0.77 and 1.63 kg. 

Prob. 10. Indicate roughly the principal paths of magnetic leakage 
in Fig. 60, and explain the influence of the leakage upon the tractive 
effort, with a small and a large air-gap. 

Prob. 11. The flux between two thin and high bus-bars, placed at 
a short distance from each other, has the general character shown in 
Fig. 61. Calculate the force per meter length that pushes the bus-bars 
apart when, during a short-circuit, the estimated current is 50 kilo- 
amperes. Ans. About 8(X) kg. per meter. 

Prob. 12. Deduce an expression for the magnetic pull due to the 
eccentric position of the armature in an electric machine (Fig. 62). 
A certain allowance is usually made for this pull in addition to the 
weight of the revolving part, in determining the safe size of the shaft. 

Solution : Since the pull is proportional 
to the square of the flux density, we 
replace the actual variable air-gap 
density by a constant radial density 
acting upon the whole periphery of the 
armature and equal to the quadratic 
average (the effective value) of the 
actual flux density distribution. Let 
this value be Bgff kl.per sq. cm. when 
the armature is properly centered. 
Let the original uniform air-gap be a, 
and the eccentricity be e. Since a and 
e are small as compared to the diameter 
of the armature, the actual air-gap 
at an angle a from the vertical is 
approximately equal to a — ecos a. Neg- 
lecting the reluctance of the iron parts 
of the machine, the flux density is inversely as the length of the air- 
gap, so that we have 

Ba=Beffa/(a-€ cos a) = Beff/[l-{£/a) cos a]. 

Let A be the total air-gap area to which B refers; then, according 




a— e 
Fig. 62. — ^An eccentric armature* 



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Chap. XIII] TORQUE AND TRACTIVE EFFORT 249 

to eq. (171), the vertical component of the pull upon the strip of the 
width da is 

(f/?'e* = (B«V24.7)(A/2;r)da.cos a. 

The horizontal component of the pull is balanced by the corresponding 
component on the otlier half of the armature. The total pull downward is 

Fex'^[2A/(27tX24.7)]C B«»cos a da. 

Putting tan a=z and integrating we get the so-called Sumec formula 
for the eccentric pull, in kg. : 

Fex«(ABe//V24.7)(e/a)[l-(€/a)»^l•^ . . . (175) 

The integration is simplified; if, before integrating, the difference is 
taken between the vertical forces at the points corresponding to 
a and to ;r — a. The limits of integration are then and i?r.* 

Prob. 13. The average flux density under the poles of a direct- 
current machine is 6.5 kl/sq.cm.; the poles cover 68 per cent of the 
periphery. The diameter of the armature is 1.52 m.; the effective 
length is 56 cm. What is the magnetic pull when the eccentricity is 
10 per cent and when it is 50 per cent of the original air-gap? 

Ans. 3.2 and 24 metric tons. 

Prob. 14. The 22/2-kv., 2500-kva., transformer specified in prob. 
5, Art. 64, had a total impedance drop of 73.5 volts at full load current 
on the low-tension side. "V^Hiat average force is exerted on each coil- 
face during a short-circuit, provided that the line voltage remains 
constant? The transformer winding consists of 12 high-tension coils of 
100 turns each, and of 11 low-tension coils interposed between the high- 
tension coils, together with 2 half-coils at the ends. Two of the dimen- 
sions of the coils are repeated here: Om»2.6 m.; Z=18 cm. Hint: The 
short-circuit current is equal to 20(X)/73.5=27.2 times the rated current. 

Ans. About 22. metric tons. 

Prob. 15. What is the mechanical pressure on the surface of the 
conductors in prob. 3? 

Ans. 47.6 and 5.8 milhgrams per sq.cm. 

71. The Determination of the Mechanical Forces by Means 
of the Principle of Virtual Displacements. In order to determine 
the mechanical force or torque between two parts of a mag- 
netic circuit the general method consists in giving these parts 
an infinitesimal relative displacement and applying the law of 

^ J. E. Sumec, Berechnung des einseitigen magnetischen Zuges bei Excen- 
trizitat, ZeUsckriJt fiir Elektrotechnik (Vienna), Vol. 22 (1904), p. 727. This 
periodical is continued now under the name of Electrotechnik und Maschinr 
enbau. 



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260 THE MAGNETIC CIRCUIT [Art. 71 

the conservation of energy to this displacement. From the 
equation so obtained the component of the force in the direction 
of the displacement can be calculated. Taking other displace- 
ments in different directions, a sufficient number of the com- 
ponents of the forces are determined to enable one to calculate 
the forces themselves. Since the forces in a given position of 
the system are perfectly definite, the result is the same no matter 
what displacements are assumed, provided that these displace- 
ments are possible, that is, consistent with the given conditions 
of the problem. Therefore displacements are selected which 
give the simplest formulse for the energies involved. We have 
had two applications of this principle in the preceding article, 
in deriving the expressions for the tension and the compression 
in the field, by giving the simple magnetic circuit the proper 
" virtual '' displacements. In applying this method, not only 
the mechanical displacement has to be specified, but also the 
electric and the magnetic conditions of the circuit, in order to 
make the energy relations entirely definite. Thus, in the pre- 
ceding article, the electromagnetic condition was <^ = const.i 

First let us take the case when the partial linkages are neg- 
ligible; then according to the third eq. (99), the stored energy is 

W,=i0^(R, (176) 

where (R is the reluctance of the circuit. Let F be the unknown 
mechanical force between two parts of the magnetic circuit at 
a distance s, and let one part of the system be given an infini- 
tesimal displacement ds. Let F be considered positive in the 
direction in which the displacement ds is positive. The mechan- 
ical work done is then equal to Fds, As in the preceding article, 
let this displacement take place with a constant flux, so that 
there is no interchange of energy between the magnetic circuit 
under consideration and the electric circuit by which it is excited. 
Then the work is done entirely at the expense of the stored energy 
of the magnetic circuit, and we have : 

Fds=dW^=-dW„ (177) 

where dW^ is the mechanical work done. The sign minus before 

^ The principle of virtual displacements is much used nowadays in the 
theory of elasticity and in the calculation of the mechanical stresses in the 
so-called statically-indeterminate engineering structures. 



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Chap. XIII] TORQUE AND TRACTIVE EFFORT 261 

dWg is necessary because the stored energy decreases. From 
eqs. (176) and (177) we get 

F=-i0^,d(R/ds (178) 

In some cases it is more convenient to express F through M 
and (P. We have 

or 

F=+.iM^d(P/ds (179) 

In the preceding formulae F is in joulecens, M is in ampere- 
turns, is in webers, (P is in henrys, and (R in ymehs. With 
other units the formulse contain an additional numerical factor. 
It is to be noted that the mechanical forces are in such a direc- 
tion that they tend to increase the permeance and decrease the 
reluctance of the circuit. This agrees with previous statements. 
See Arts. 41 and 57. 

If the partial linkages are of importance, it is convenient 
to express the stored energy in the form Ws=i'PL, because the 
inductance L takes account of the partial linkages; see eqs. 
(105) and (106), Art. 58. The energy equation, according to 
eq. (177), is then 

Fds= -dTF.= -idi'PL) (180) 

and the condition that there is no interchange of energy with 
the line is 

d(iL) = (181) 

The latter equation becomes clear by reference to eq. (106a), 
because Li=n0eQ, where 0eq is the equivalent flux under the 
supposition of no partial linkages. The condition that there 
shall be no e.m.f. induced in the winding during the displace- 
ment, is d(n0eq)/dt==O, whence eq. (181) follows directly. 

Performing the differentiations in eqs. (180) and (181), and 
substituting the value of Ldi from the second equation into the 
first, we get that 

F^+iiML/ds . (182) 

When there are no partial linkages, L=n^(P, and eq. (182) 
becomes identical with (179). 



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262 THE MAGNETIC CIRCUIT [Art. 71 

FormulcB for the average force of direct currerU electromagnets. 
In a tractive magnet (Fig. 60) or a rotary magnet (Fig. 63) 
it is often required to know the average pull over a finite travel 
of the moving part. For the average force, the equation 
Faw^SsJTFmss — -4TFa holds, which is analogous to eq. (177) ; 
the only difference being that finite instead of infinitesimal incre- 
ments are used. If the motion takes place at a constant fiux, 
or at least the values of the flux are the same in the initial and 
the end-positions, we get from the preceding formulae: 

Fave^h<fH(Rl-(fi2)/{S2-Si); . . . (183) 

-i(Mi^(Pi-Mj^(p2)/(s2-Si); . . (184) 

= i(n^^i-i22L2)/(s2-«i). . . . (185) 

The finite travel of the plunger often takes place at a con- 
stant current, for instance, in the regulating mechanism of a 




Fig. 63. — ^A rotary electromagnet. 

series arc-lamp, also approximately in a direct-current electro- 
magnet connected across a constant-potential line. Under such 
conditions the foregoing formulae are not directly appliahle, 
because they have been deduced under the assumption of no 
interchange of energy between the electric and the magnetic 
circuits, so that this case has to be considered separately. 

Let the motion be in the direction of the magnetic attraction, 
and let the current remain constant during the motion. The 
stored energy is larger in the end-position than in the initial 
position, because the flux is larger and the current is the same. 
Therefore, the energy supplied during the motion from the line 
must be sufiicient to perform the mechanical work, and to 



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Chap. XIII] TORQUE AND TRACTIVE EFFORT 253 

increase the energy stored in the magnetic circuit. The increase 
in the stored energy is 

and the energy supplied from the line is calculated as follows: 
The average voltage induced during the motion of the plunger 
is eave='i{L2—Li)/t, where t is the duration of the motion. The 
energy supplied from the line is therefore 

Wi = iCa vet = i^ (L2 -Li) . 

Thus, the energy supplied from the line is twice as large as the 
work performed, and we have the following important law 
(due to Lord Kelvin) : When in a singly excited magnetic circuit, 
without saturationj a deformation takes place, at a constant current, 
the energy supplied from the line is divided into two equal parts, 
one half increasing the stored energy of the circuit, the other half 
being converted into mechanical work. 

According to this law we have, for a constant current electro- 
magnet, that the mechanical work done is equal to the increase 
in the energy stored in the magnetic field. Hence 

Fane^S^JWm^+JW,. 

Thus, 

Fave-mL2-Li)/(s2'-Si); (186) 

or, if the partial linkages are negligible, 

Fave==i{^2^(R2-^i'(Rl)/{S2'-Si)', . . . (187) 

-^iMH(P2-(Pi)/(s2'Si) (188) 

When a magnet performs a rotary motion (Fig. 63), the 
preceding formulse are modified by substituting Td6 in place of 
Fds, or Tave(02-'0i) in place of Fave(s2—si), Here T is the torque 
in joules and (^2—^1) or dO is the angular displacement of the 
armature in radians. Or else, in the foregoing formulae F may 
be understood to stand for the tangential force, and the dis- 
placement to be ds=r dO, where r is the radius upon which 
the force F is acting. Then the torque is r= Fr. If, for instance, 
we apply eq. (179) to a rotary motion, it becomes 

T:=Fr^+iM^d(P/d0 (189) 

The other equations may be written by analogy with this one. 



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254 THE MAGNETIC CIRCUIT [Art. 71 

Alternating-Current Electromagnets, The preceding formulse are 
deduced under the supposition that the magnetic field is excited 
by a direct current. They are, however, applicable also to alter- 
nating-current electromagnets, because the pulsations in the 
current merely cause the energy to surge to and from the magnetic 
circuit, without any net efifect, so far as the average stored energy 
and the mechanical work are concerned. The average stored 
energy corresponds to the effective values of the current and the 
flux. 

In practice two types of alternating-current electromagnets 
are of importance, namely, those operating at a constant voltage 
and those operating at a constant current. As an example of 
the first class may be mentioned the electromagnets used for 
the operation of large switches at a distance (remote control); 
the windings of such electromagnets are usually connected directly 
across the line. Constant-current magnets are used in the 
operating mechanism of alternating-current series arc-lamps. In 
an A. G. electromagnet practically aU of the voltage drop is 
reactive and hence proportional to the flux. 

In a constant-potential electromagnet the effective value of 
the equivalent flux is the same for all positions of the plunger 
(neglecting the ohmic drop in the winding). Therefore, formula 
(185) holds true. Let e be the effective value of the constant 
voltage, or more accurately the reactive component alone, and 
let /be the frequency of the supply. Then, e=27:fLiii = 2nfL<^2f 
so that the formula for the pull becomes 

Fave=^e(ii^l2)/[^7tf{S2-Si)] (190) 

For a constant-current A.C. electromagnet eq. (18® applies; 
introducing again the reactive volts ei = 27r/Lii and e2=2;r/L2i, 
we get 

Fave=i{e2-ei)/[^f(s2-8i)] (191) 

In both cases the mechanical work performed is proportional 
to the difference in the reactive volt-amperes consumed in the 
two extreme positions of the moving part.^ 

^ For further details in regard to electromagnets consult C. P. Steinmetz, 
Mechanical Forces in Magnetic Fields, Trans, Amer, Inst. Elec. Engs., Vol. 
30 (1911), and the discussion following this paper; also C. R. Underbill 
Solenoids, Electromagnets, and Electromagnetic Windings{\^Wi), chapters 6 



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Chap. XIII] TORQUE AND TRACTIVE EFFORT 255 

Prob. 16. Derive formula (171) for the lifting magnet by means 
of the principle of virtual displacements. Solution: The reluctance 
of the air-gap is (R=s/(/itA), so that cKR/cte = l/(/iii), hence, according 
to eq. (178) F= -i0V(M) = -ABy2pL. The minus sign indicates that 
the stress is one of tension. 

Prob. 17. Derive expression (172) from formula (179). 

Prob. 18. Derive the formula for the repulsion between the wingings 
in a transformer from eq. (182). 

Prob. 19. Derive from eq. (182) the force of repulsion between two 
infinitely long, parallel, cylindrical conductors placed at a distance 
of b meters apart, and forming an electric circuit (Fig. 47). 

Ans. 2MiHl/b) XlO'^ kg. for I meters of the loop. 

Prob. 20. What deformation of the windings may be expected 
diuing a severe short-circuit of a core-type or a cruciform type trans- 
former (Figs. 12 and 14) with cylindrical coils, (a) when the centers 
of the coils are on the same horizontal line, and (6) when one of the 
windings is mounted somewhat higher than the other ? 

Prob. 21. Show that in a constant-current rotary magnet (Fig. 
63) ra= Const. — that is, the torque in the different positions of the 
armature is inversely proportional to the air-gap at the entering pole- 
tip. Hint : cKP = /iwrdd/a, where w is the dimension . parallel to the 
shaft. 

Prob. 22. State Kelvin's law when mechanical work is done against 
the forces of the magnetic field. 

Prob. 23. A 60-cycle, 8-amp., series arc-lamp magnet has a stroke 
of 32 mm.; the reactive voltage consumed Jn the initial position is 9 
v., and in the final position 20 v. What is the average pull ? 

Ans. 372 grams. 

72. The Torque in Generators and Motors. The magnetic 
circuits considered in the preceding articles of this chapter are 
singly excited, that is, they have but one exciting electric circuit. 
From the point of view of mechanical forces this also applies 
to each ?.ir-gap in a transformer, because, neglecting the mag- 
netizing current, the primary and the secondary coils may be 
combined into equivalent leakage coils (Art. 64). On the other 
hand, a generator or a motor under load has a doubly-excited 
magnetic circuit, the useful field being linked with both the 
field and the armature windings. The two m.m.fs. not being 
in direct opposition in space, the flux is deflected from the shortest 

to 9 incl.; S. P. Thompson, On the Predetermination of Plunger Electro- 
magnets, Intern, Elect. Congress, St. Louis, 1904, Vol. 1, p. 542; E. Jasse, 
Ueber Elektromagnete, Elecktrotechnik und Maschineribau, Vol. 28 (1910), 
p. 833; R. Wikander, The Economical Design of Direct-current Electro- 
magnets; Trans. Amer. Inst. Elect. Engs., Vol. 30 (1911). 



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256 



THE MAGNETIC CIRCUIT 



[Abt. 72 



path, and the torque is due to the tendency of the tubes of force 
to shorten themselves longitudinally, and to spread laterally. 

Consider the simplest generator or motor, consisting of a 
very long straight conductor which can move at right angles 
to a uniform magnetic field of a density B (Fig. 64). Let the 
ends of the conductor slide upon two stationary bars through 
which the current is conducted into a load circuit in the case of 
a generator action, and through ^which the power is supplied 
in the case of a motor action. If the magnetic circuit contains 

no iron, tl^e resultant field is a 
superposition of the original uniform 
field and of the circular field created 
by the current in the conductor. 
With the direction of the current 
indicated in Fig. 64, the resultant 
field is stronger on the left-hand side 
of the conductor than it is on the 
right-hand side, and there is a 
resultant lateral pressure exerted 
upon the conductor to the right. 
In the case of a motor action the 
direction of the motion is in the 
same direction as the pressure from 
the stronger field. In the case of a 
generator action the conductor is 
moved by an external force against this pressure. Compare 
also the rule given in Art. 24. 

To find the mechanical force between the conductor and the 
field, we will apply again the principle of virtual displacements. 
Let the current through the conductor be i, and let the con- 
ductor be moved against the magnetic pressure by a small amount 
ds. Assume that the stored magnetic energy of the electric 
circuit to which the conductor belongs is the same in the various 
positions of the conductor. (Such is the case in actual machines.) 
Then the work done by the external force is entirely converted 
into electrical energy, and we have 




Fig. 64. — ^A straight conductor 
in a uniform magnetic field. 



Fds=iedt, (192) 

where e is the induced e.m.f. Let both F and e refer to a length 



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Chap. XIII] TORQUE AND TRACTIVE EFFORT 267 

I of the conductor. Substituting the value of e from eq. (27), 
Art. 24, we have 

Fds/dt=iBlv, 
or, since ds/dt=v, 

F=iBl (193) 

In this expression i is in amperes, J5 is in webers per sq.cm., 
/ is in cm., and F is in joulecens. With other imits the formula 
contains a numerical factor. 

Formula (193) maybe used also with anon-uniform field, and 
also when the direction of the conductor is not at right angles 
to that of the line of force. In such cases the formula becomes 
dF^^iBdlj where dF is the force acting upon an infinitesimal 
length dl of the conductor, and B is understood to be the com- 
ponent of the actual flux density perpendicular to dl. 

As an application of formula (193), consider the attraction 
or the repulsion between two straight parallel conductors carrying 
currents ii and 12, and placed at a distance b from each other. 
The circuit of each conductor may be considered closed through 
a concentric cylindrical shell of infinite radius, as in Art. 60. 
It is apparent from symmetry that the field produced by each 
system gives no resultant force with the current in the same 
system. Thus, the mechanical force is due to the action of the 
field 1 upon the current 2, and vice versa. 

The flux density due to the system 2 at a distance b from 
the conductor 2 is -B=//.t2/2;r6, so that, according to eq. (193), 

F=/£iii2Z/27r6 joulecens, (194) 

where /£= 1.257X 10"^. In kilograms the same formula is 

2^=2.04iii2(Z/6)10-8, (194o) 

provided that I and b are measured in the same imits. The 
force is an attraction or a repulsion according to whether the 
two currents are flowing in the same or in the opposite directions. 
When ii = i2, this formula checks with that given in prob. 19 
above. 

Formula (193) applies also to the tangential force between 
the field and armature conductor in any ordinary generator or 
motor, provided that (a) the conductors are placed upon a 
smooth-body armature, and (6) the conductors are distributed 



/. ^ 



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258 THE MAGNETIC CIRCUIT [Art. 72 

uniformly over the armature periphery, so that the stored mag- 
netic energy is the same in all positions of the armature. It 
would be entirely wrong, however, to apply this formula to 
a slotted armature, using for B the actual small flux density 
in the slot within which the conductor lies. This would give 
the force acting upon the conductor itself, and tending to press 
it against the adjacent conductor or against the side of the slot; 
but the actual tangential force exerted upon the armature as 
a whole is many times greater, and practically all of it is exerted 
directly upon the steel laminations of the teeth. 

At no-load, the flux distribution in the active layer of the 
machine is symmetrical with respect to the center line of each 
pole (Fig. 24), so that the resultant pull along the lines of force 
is directed radially. The armature currents distort the field 
as a whole, and also distort it locally around each tooth, the 
general character of distortion being shown in Fig. 36. The 
unbalanced pull along the lines of force has a tangential com- 
ponent which produces the armature torque. This torque, 
although caused by the current in the armature conductors, 
is largely exerted directly upon the teeth, because the flux density 
there is much higher. 

Thus, in order to determine the total electromagnetic torque 
in a slotted armature, it is again necessary to apply the principle 
of virtual displacements. The reluctance of the active layer 
per pole varies somewhat with the position of the armature, 
so that the energy stored in the field is also slightly fluctuating. 
It is convenient, therefore, to take a displacement which s a 
multiple of the tooth pitch, in order to have the same stored 
energy in the two extreme positions. This gives the average 
electromagnetic torque. 

(a) The Torque in a Direct-imrrent Machine. Let the virtual 
displacement be equal to geometric degrees and be accomplished 
in t seconds. Then we have 

Ta^e=iEi, (195) 

where T is the torque, i is the total armature current, and E is 
the total induced e.m.f. Eq. (195) states the equality of the 
mechanical work done and of the corresponding electrical energy 
supplied. The average induced eim.f. is independent of the 
flux distribution, or of the presence or absence of teeth (see 



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Chap. XIIII .TORQUE AND TRACTIVE EFFORT 259 

Art. 24 and prob. 18 in Art. 26). Take to correspond to 
two pole pitches, or 0=2n/{^'p)\ then t=l/fi where / is the 
frequency of the magnetic cycles. Substituting these values and 
using the value of E from eq. (37), Art. 31, we get, after 
reduction, 

Tave^iNv^/n\ovleB, (196) 

where is in webers; or 

raw=0.0325iiVp<^XlO-2kg-meters, . . (196a) 

being in megalines. 

This formula does not contain the speed of the machine, 
the torque depending only upon the armature ampere-turns iN 
and the total flux p0. Consequently, the formula can be used 
for calculating the starting torque or the starting current of 
a motor. Eqs. (196) and (196a) give the total electromagnetic 
torque, part of which serves to overcome the hysteresis, eddy 
currents, friction and windage. The remainder is available on 
the shaft. When calculating the starting torque, it is necessary 
to take into account the effort required for accelerating the 
revolving masses. 

(ft) The Torque in a Synchronous Machine. The equation of 
energy is 

TaveO=-niiE COB (l>\t, (197) 

where m is the number of phases, i and E are the effective values 
of the armature current and the induced voltage per phase, and 
(f>' is the internal phase angle (Fig. 37). Taking again a dis- 
placement over two poles and using the value of E from eq. (3) 
Art. 26, we get 

Tave=0.0361ki,miN cos 0'p<^lO"2 kg-meters. . (198) 

(c) The Torque in an Induction Machine. The torque being 
exerted between the primary and the secondary members of 
an induction machine, it may be considered from the point of 
view of either member. This is because the torque of reaction 
upon the stator is equal and opposite to the direct torque upon 
the rotor. For purposes of computation it is more convenient 
to consider the torque from the point of view of the primary 
winding, in order to be able to use the primary frequency and 



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260 THE MAGNETIC CIRCUIT [Abt. 72 

the synchronous speed. Therefore eq. (198) gives the torque of 
an induction machine (including, as before, friction and hysteresis), 
where the various quantities refer to the stator. However, these 
quantities may equally well be taken in the secondary, but in 
this case, since hysteresis occurs mainly in the stator, the torque 
to overcome hysteresis is not included. 

Formula (198) is hardly ever used in practice, especially 
for the computation of the starting torque, because it is difficult 
to eliminate the large leakage flux which gives no torque. It 
is much more convenient to determine the torque from the circle 
diagram, or from the equivalent electric circuit. 

In case the torque is determined from the equivalent electric cir- 
cuit, we can write from eqs. (196) and (197) the expressions for the 
torque directly, by substituting for d its value 21c•^(R.P.M.)/60. 
For a direct-current machine, 

r«=0.0326— — -— TT-rKfr-metere. . . . (199) 

X • (xv.ir .M.^ 

Here the induced e.m.f. E^EtdtiRa, according to whether the 
machine is a generator or a motor; Et is the terminal voltage 
and Ra the resistance of the armature, brushes, and series field. 
For an alternating-current machine, 

rn ^ ^«^.30mi^cos A' ^ ,^^^v 

Taw = 0.0325— — — -7t-Kg.-meters. . . . (200) 
x(R.P.M.) 

In a Sjmchronous machine E is the induced e.m.f. and 0' is 
the internal phase angle. In an induction machine, iE cos ^' 
is the power per phase delivered to the rotor, that is, the input 
minus hysteresis and primary PR loss. The term (R.P.M.) is 
in all cases the s3aichronous speed of the machine. 

Prob. 24. Two single-conductor cables from a direct-current machine 
are installed parallel to each other at a distance of 16 cm. between their 
centers, on transverse supports spaced 80 cm. apart. The rated current 
through the cables is 850 amp. What is the force acting upon each 
support under the normal conditions, and when the cturent rises to 
twenty times its rated value during a short-circuit? 

Ans. 0.0737 and 29.5 kg. 

Prob. 25. A 4-pole, series direct-cturent motor must develop a 
starting torque of 74 kg.-m. (including the losses). The largest possible 
flux per pole is about 2.5 ml.; there are 240 turns in series between the 
brushes. Calculate the starting current. Ans. 95 amp. 



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Chap. XIII] TORQUE AND TRA.CTIVE EFFORT 261 

Prob. 26. Explain the reason for which the compensating winding 
(Art. 54), while removing the armature reaction, does not affect appre- 
ciably the useful torque of the motor. Hint: this can be shown by 
applying the method of virtual displacements; also from the fact that 
the local distortion of the flux in the teeth is not removed. 

Prob. 27. On the basis of Art. 15, explain the mechanism by which 
the phenomenon of hysteresis in an armature core causes an opposing 
torque. Explain the same for eddy currents. 

Prob. 28. Demonstrate that formula (193) may be used for slotted 
armatures, provided that B stands for the average flux density per 
tooth pitch. Hint: take a virtual displacement of one tooth pitch, 
and express the induced voltage through the flux J0^BIX per tooth 
pitch. 

Prob. 29. Show that in a single-phase synchronous motor the 
torque at the synchronous speed pulsates at double the frequency of 
the supply; also that the torque is zero at any but the synchronous 
speed. 

Prob. 30. Prove that in an induction machine the torque near syn- 
chronism is approximately proportional to the square of the voltage 
and to the per cent slip. Hint: t'—CUonst. X Xslip. 

Prob. 31. Describe in detail how to calculate the maximum starting 
torque of a given induction motor, and how to calculate th^ amount 
of secondary resistance necessary for a prescribed torque. 



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APPENDIX I 
THE AMPERE-OHM SYSTEM OF UNITS 

The ampere and the ohm can be now considered as two 
arbitrary fundamental units established by an international 
agreement. Their values can be reproduced to a fraction of a 
per cent according to detailed specifications adopted by practically 
all civilized nations. These two units, together with the centi- 
meter and the second, permit the determination of the values of 
all other electric and magnetic quantities. The units of mass and 
of temperature do not enter explicitly into the formulae, but are 
contained in the legal definition of the ampere and of the ohm. 
The dimension of resistance can be expressed through those of 
power and current, according to the equation P=PRj but it 
is more convenient to consider the dimension of 72 as fundamental 
in order to avoid the explicit use of the dimension of mass [M]. 

For the engineer there is no more a need of using the electro- 
static or the electromagnetic units; for him there is but one 
ampere-ohm system^, which is neither electrostatic nor electro- 
magnetic. The &mpere has npt only a magnitude, but a physical 
dimension as well, a dimension which with our present knowledge 
is fundamental, that is, it cannot be reduced to a combination 
of the dimensions of length, time, and mass (or energy). Let 
the dimeiisions of current be denoted by [I] and that of resistance 
by [R] ; let the dimensions of length and time be denoted by the 
commonly recognized symbols [L] and [T]. The magnitudes and 
the dimensions of the important magnetic units are expressed 
through these four, as is shown in the following table. For the 
expressions of the electric and the electrostatic quantities in the 
ampere-ohm system see the author's " Electric CircuiV 

Other units of more convenient magnitude are easily created 
by multiplying the above-tabulated units by powers of 10, or 
by adding prefixes milli-, micro-, kilo-, mega-, etc. 

A study of the physical dimensions of the magnetic quantities 

263 



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AMPERE^HM SYSTEM 



263 



is interesting in itself, and gives a better insight into the nature 
of these quantities. Moreover, formulsB can sometimes be checked 
by comparing the physical dimensions on both sides of the equa- 
tion. Let, for instance, a formula for energy be given 

where fc is a numerical coefficient. Substituting the physical 
dimensions of all the quantities on the right-hand side of the 
equation from the table below, the result will be found to have 
the dimension of energy. This fact adds to one's assurance 
that the given formula is theoretically correct. 

TABLE OF MAGNETIC UNITS AND THEIR DIMENSIONS IN 
THE AMPERE-OHM SYSTEM 



Symbol and Formula 


Quantity. 


Dimension. 


Name of the Unit. 


M^nl 


Magnetomotive force 


[I] 


Ampere-tiun. 


H^M/l 


Field intensity, or 


[IL-^] 


Ampere-turn per 




m.m.f. gradient 




centimeter. 


0=ET 


Magnetic flux 


[IRT]* 


Weber (maxwell). 


B^0IA 


Magnetic flux density 


[IRTL-T 


Webers (maxwells) 
per square centi- 
meter. 


(P^0/M 


Permeance 


[RT] 


Henry (perm). 


(R^M/9^l/(P 


Reluctance 


[R-^T-*1 


Ymeh (rel). 
Henries (perms) per 


v=B/H 


Permeability 


[RTL-»] 


centimeter cube. 


v^H/B = \/ti 


Reluctivity 


[R-*T-*L] 


Ymehs (rels) per 
centimeter cube. 


w=w^ 


Magnetic energy or 


[I'RT] 


Joule or watt-sec- 




work 




ond. 


W/v=^\BH 


Density of magnetic 


[I'RTL-'] 


Joules per. cubic 




energy 




centimeter. 


F^W/l 


Force 


I'RTL-T 


Joulecen. 



* This is also the dimension of the magnetic pole strength. The concept 
of pole strength is of no use in electrical engineering, and, in the author's 
opinion, its usefulness in physics is more than doubtful. The whole theory 
of electromagnetic phenomena can and ought to be built up on the two 
laws of circuitation, as has been done by Oliver Heaviside in his Electro- 
magnetic Theory, 

A small irregularity is due to the use of the maxwell and of 
its multiples instead of the weber. As long as this usage persists 



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264 THE MAGNETIC CIRCUIT 

it is convenient to use the corresponding units for reluctance 
and permeance, to which the author has ventured to give the 
names of rel and perm. Since one maxwell is equal to 1/10* of a 
weber, one perm is equal to 1/10* of one henry, and one rel is 
10* ymehs. Accordingly, permeabilities and reluctivities are 
measured in perms per centimeter cube and in rels per centimeter 
cube respectively. 

In order not to break with the established usage, the maxwell, 
the perm, and their multiples are employed in numerical compu- 
tations in this book, while the weber and the henry are used in 
the deduction of the formulae, being the natural fundamental 
units of flux and permeance in the ampere-ohm system. It is pos- 
sible that the constant necessity for multiplying or dividing results 
by 10""*, due to the use of the maxwell, may prove to be more and 
more of an inconvenience in proportion as magnetic computations 
come into common engineering practice. Then the weber, the 
henry, and their submultiples will be found ready for use, and 
the system of magnetic units will be completely coordinated. 

Another irregularity in the system as outlined above is caused 
by the use of the kilogram as the unit of force, because it leads to 
two units for energy and torque, viz., the kilogram-meter and 
the joule; 1 kg.-meter= 9.806 joules. Force ought to he measured 
in joules per centimeter length, to avoid the odd multiplier. Such 
a imit is equal to about 10.2 kg., and could be properly called 
the jouLecen (= 10^ dynes). There is not much prospect in sight 
of introducing this unit of force into practice, because the kilo- 
gram is too well established in common use. The next best 
thing to do is to derive formulae and perform calculations, whenever 
convenient, in joulecens, and to convert the result into kilo- 
grams by multiplying it by gr== 9.806. This is done in some 
places in this book. 

Thus, leaving aside all historical precedents and justifications, 
the whole system of electric and magnetic units is reduced to 
this simple scheme : In addition to the centimeter, the gram, the 
second and the degree Centigrade, two other fundamental imits 
are recognized, the ohm and the ampere. All other electric and 
magnetic units have dimensions and values which are con- 
nected with those of the fundamental six in a simple and almost 
self-evident manner (see the table above). 

To appreciate fully the advantages of the practical ampere- 



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AMPERE^HM SYSTEM 



265 



ohm system over the C.G.S. electrostatic and electromagnetic 
systems, one has only to compare the dimensions, for instance, 
of magnetomotive force and of flux in these three systems, 
as shown below. 





The Ampere- 
Olim System. 


C.G.S. Electro- 
magnetic System. 


C.O.S. Electfo- 
statie System. 


Dimension of m.m.f . 
Dimension of flux. 


m 

[IRT] 


L»M»T-V-» 


LiMiK-i 



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APPENDIX II 
AMPERE-TURN vs. GILBERT 

The reader has probably been taught before that the per- 
meability of air is equal to unity in the electromagnetic C.G.S. 
system; silent assumption was then probably made that /£==! 
also in the practical ampere-ohm system. The true situation 
is, however, as follows: In any system of units whatsoever, the 
fundamental equation tf=(/£A/Z).Af holds true, being a mathe- 
matical expression of an observed fact. Now let the quantities 
be expressed in the ampere-ohm system, and assmne the centi- 
meter to be the imit of length. The flux is then expressed 
either in maxwells or in webers, both of which are connected 
with the ampere-ohm system through the volt. The natural 
(though not the only possible) unit for the magnetomotive force 
is one ampere-turn. Therefore, all the quantities in the fore- 
going equation are determinate, *and the value of /i cannot be 
prescribed or assumed, but must be determined from an actual 
experiment, the same as the electric conductivity of a metal, 
or the permittivity of a dielectric have to be determined. 
Experiment shows that /£= 1.257 when the maxwell is used as the 
imit of flux, and hence /£= 1.257 X 10~® if the flux is measured in 
webers. 

It is possible to assume /£= 1, provided that the imit of mag- 
netomotive force is not prescribed in advance. In this case, 
the imit of magnetomotive force, as determined from experiment, 
comes out equal to 1/1.257 of an ampere-turn. This unit is 
called the gilbert, and it must be understood that the permeability 
of non-magnetic materials is equal to unity only if the magneto- 
motive force is measured in gilberts. To the author the advan- 
tages of such a system for practical use are more than doubtful. 
In the first place, the gilbert is a superfluous imit, because the 
results of calculations must after all for practical purposes be 
converted into ampere-turns in order to specify the number of 

266 



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AMPERE-TURN vs. GILBERT 267 

turns and the exciting current of windings. Thus, one would 
have to deal with two units of magnetomotive force, the gilbert 
and the ampere-tum, one being about 0.8 of the other. In the 
second place, with the assumption /£= 1 for non-magnetic materials 
B becomes numerically equal to H, which is a grave inconvenience, . 
because B and H are different physical quantities. B and H 
have different physical dimensions, because /£ has a definite 
physical dimension, even though the numerical value of it is 
assumed to be equal to unity for air. Therefore, to be sure 
that proper physical dimensions are preserved, one has to remem- 
ber where /£ is omitted in formulae, and for a physical interpretation 
of results it is much more convenient to have it there, explicitly. 

Still another objection to using the gilbert and to putting 
jj. equal to unity for air is that the ratio of the ampere-tum 
to the gilbert is equal to a quasi-scientific constant 4;r/10. 
To the author's knowledge, there is no simple, elementary way 
of deducing the value of this constant, without going over 
the whole mathematical theory of electricity and magnetism. 
Thus, a constant is retained in practical formulae, the significance 
of which remains a puzzle to the engineer aU his life. It is true 
that the value of /£= 1.257 is equal to the same 4;r/10 after all; 
but in this case there is nothing ''absolute,'' mysterious, or 
sacred about the value of 4;r/10. The student is simply told 
that 1.257 happens to be equal to 4;r/10 because the value of 
the ampere was unfortunately so selected. It is not necessary 
to go into further details, because the historical reasons which 
led to the selection of the values of unit pole and unit current 
hardly hold at present. All calculations would be just as con- 
venient if fi were equal to 2.257, or any other value, instead of 
1.257. 

For these reasons the author unhesitatingly discards the 
gilbert in teaching as well as in practice and uses the ampere- 
tum as the natural unit of magnetomotive force.' The value 
of permeability becomes then an experimental quantity which 
depends upon the units selected for flux and length. 



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INDEX 



PAGB 

Active layer characteristic, definition of 168 

Aging of laminations 49 

Air-gap ampere-turns 89 

factor, definition of 90 

factor for induction machines 98 

flux, nature of its distribution 88 

permeance and m.m.f ., accurate method 92-97 

simplified, permeance of 89 

Alternating current electromagnets, tractive effort of 254 

machines, torques of 259-260 

machines. See also Synchronous and Induction. 
Alternator. See Synchronous machine. 

calculation of regulation 155 

definition of load characteristic 141 

definition of regulation 156 

wave-form 72 

Ampere-Ohm system 262 

Ampere-turns, the cause of magnetism. See also M.m.f 4 

Ampere-turn, 

unit of m.m.f 5 

V8, Gilbert 266 

Analogue, mechanical, to hjrsteresis 36 

to inductance 185 

Annealing of laminations 50 

Apparent flux density in teeth 101 

Armature, eccentric, force upon 248 

reactance in synchronous machine, nature and definition of 140 

See also Inductance. 

reaction, definition of , 140 

in a direct current machine 163 

in an induction machine 131 

in a rotary converter 175 

in a synchronous machine 139 

Asynchronous. See Induction machines. 
Auxiliary poles. See Commutating poles. 

Axial length of armature, effective 94 

269 



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270 INDEX 

PAQB 

Back ampere turns. See Demagnetizing, Direct. 

Belt leakage, description of 223 

Belts of current in a D.C. machine 163 

B-H curves for iron 20-24 

relation for air 15 

Blondel diagram 154-155 

Breadth factor, definition of 65 

formula for 68 

Brush shift in D.C. machines 165 

Bus-bars, repulsion between 248 

Cable, concentric, flux distribution in 189 

induction of a single phase 191 

Carter's curve for fringe permeance 95 

Castings 21 

Characteristic, active layer, definition of 168 

air 231 

load, definition of in alternator' 141 

open circuit. See Saturation curves, and Exciting cur- 
rent. 

phase, synchronous motor 142 

Circuit, a simple magnetic. (See also Magnetic circuit) 1 

magnetic, containing iron, series-paraUel 29 

definition of 3 

Coercive force 32 

Commutating poles, calculations for. (See also Interpoles) 173 

definition of and location 172 

Commutation, criterion of 235 

description of 233 

frequency of 236 

of a fractional pitch winding 238 

of a lap winding 235 

of a multiplex winding 238 

of a two circuit winding ^7 

vs. inductance 232 

Compensating winding for D.C. machine 174 

Complete linkages, definition of 181 

Compounding in a D.C. machine 170 

Compression, lateral, in a magnetic field 244 

Computations for interpoles 173 

Condenser, synchronous 157 

Conductor, force on, in a magnetic field 256 

Cores of iron wire 41 

of revolving machinery, m.m.f . for 105 

of transformers, m.mi for 81 

Core loss current, in an induction motor. See Core loss in revolving ' 
machinery. 



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INDEX 271 

PAGE 

Core loss current, in a transformer 82 

curves 45 

extrapolation of 64 

in transformers 82 

in revolving machinery 46 

measurement of 44 

or iron loss, definition of 42 

Cross ampere-turns. See Reaction. 

Current. See also Exciting current. 

belts in a D.C. machine 163 

secondary, in induction machine 132 

Demagnetization and distortion. See Reaction. 

Demagnetizing reaction, in direct current machines 164 

in s3mchronous machines. See Direct reaction. 

Density of flux, definition of 14 

in t^eth, apparent 101 

D^ri winding 174 

Dimensions of units, table of • 263 

Direct current electromagnet, average tractive effort of 252 

Direct current machine, brush shift 165, 167 

compensating windings for 174 

compounding in 170 

ciurent belts in 163 

demagnetizing reaction in 164 

e.m.f . induced in 75 

interpoles in 172 

inductance of coils in 236 

loaded, field m.m.f 170 

loaded, flux distribution 168 

loaded, m.inf . to overcome distortion 169 

torque 258-260 

transverse reaction in 165 

Direct current machines. See also Saturation curves. 

windings, vs. induced e.m.f 76 

vs. inductance 237 

Direct current motor, speed under load 171 

Direct reaction. See also Demagnetizing reaction. 

calculation of coefficient of 158 

definition of coefficient of 153 

nature of 150 

Displacements, virtual, principle of 249 

Distortion and demagnetization. See Reaction. 

of field in a direct current machine, m.m.f. needed to over- 
come 169 

Distributed windings, advantages and disadvantages of 66 

definition of 121 



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272 INDEX 

PAGE 

Distributed windings, for alternator fields 121 

e.m.f. of. See E.m.f. 

m.m.f. of D.C 165 

m.m.f . of single-phase 123 

m.mi . of polyphase 128 

Eccentric armature, force upon 248 

Eddy currents, nature of 40 

prevention of 41 

separation of, from hysteresis 53 

Electric circuit, equivalent, torque from 260 

Electric loading, specific, definition of 163 

Electromagnet, A.C., average tractive effort of 254 

average tractive effort of D.C 252 

lifting 243 

rotary, torque of 253 

tractive, force of 248 

Electromagnetic inertia 180 

E.m.f., average reactance, as criterion for conmiutation 235 

formula for induced 57 

induced in A.C. machines 65 

induced in D.C. machines 75 

induced in an unsymmetrically spaced transmission line 205 

methods of inducing 55 

ratio. (See Ratio of transformation.) 

ratio in a rotary converter 78 

regulation in an alternator 156 

regulation in a rotary converter 78 

vs. inductance 185 

Energy, density of, in magnetic field 240 

lost in hysteresis cycle 38 

of magnetic field, none consumed in 177 

stored in magnetic field, description of 177 

stored in magnetic field, formulae for 181 

Equivalent electric circuit, torque from 260 

permeance, definition of 184 

secondary winding, reduced to primary 133 

Exciting current in an induction motor 130 

in a transformer vs. magnetizing current 80 

of a transformer, exciting volt-amperes 82 

of a transformer, saturated core 84 

unsaturated core 81 

of machinery. See Saturation curves. 

Factor, air-gap, definition of 90 

in induction motor 98 

amplitude 84 



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INDEX 273 

PAGE 

Factor, breadth factor 65 

leakage, calculation of 108 

leakage, definition of 107 

slot factor 69 

space factor in iron 41 

winding pitch factor 71 

Faraday's law of induction 57 

Fictitious poles as assumed in synchronous machines 151 

Field frames, m.m.f. for 106 

Field, ma^etic (See also Magnetic field) 1 

m.mi. at no load. See Saturation curves. 

m.m.f. in loaded D.C. machine 170 

loaded synchronous machines 148, 156 

Field pole leakage, calculation of 110 

effect of 108 

effect of load upon 113 

effect of saturation upon 112 

Field poles, m.m.f. for 107 

Figure of loss 47 

Fleming's rule 59 

Flux. See also Leakage flux. 

air-gap, nature of its distribution 88 

density, definition of 14 

in teeth, apparent 101 

description of 5 

distribution of, in a concentric cable 189 

in a loaded D.C. machine 169 

in a loaded synchronous machine 139 

distortion. See Reaction . 

fringing, definition of 88 

permeance of 93 

gliding or revolving 126 

leakage, definition of 16 

in induction machine, description of 221 

in revolving machinery, nature of 86 

refraction 119 

units of 6 

Force, lines of magnetic 2 

mechanical, in magnetic field, average tractive effort in D.C. 

electromagnets 252 

formulae for the actual force .... 251 

lateral compression . . . , 244 

longitudinal tension 242 

of A.C. magnets 254 

of a lifting magnet 243 

of a tractive magnet 248 

on an eccentric armature 248 



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274 INDEX 

PAGB 

Force, mechanical, in magnetic field, on conductors carrying current . . 256 

on transformer coils 245 

pinch phenomenon 246 

reason for 242 

repulsion between bu»>bar8 248 

torque in a rotary magnet 253 

Fourier, method for analyzing waves for their harmonics, used 124, 161 

Fractional pitch windings, advantages and disadvantages of 67 

effect on inductance in induction machines . . 225 

effect on inductance in synchronous machines 231 

V8, commutation in direct current machines. . . 238 

Frequency of commutation 236 

Fringing, definition of 88 

Full load m.m.f . in a D.C. machine 170 

in an induction machine 131 

in a synchronous machine 148, 156 

Gauss 14 

Generator action 66 

Gilbert 12, 266 

Gliding and pulsating m.m.f 126 

Harmonics, of e.m.f ., in alternating current machines 72 

V8. voltage ratio in a rotary coverter 79 

of rectangular m.m.f . wave 123 

upper, of m.m.f. in an induction machine 136 

Heart, definition of 215 

Heating due to hjrsteresis 38 

Henry as unit of inductance 184 

permeance, definition of 9 

Herring's experiment , 56 

Hysteresis and saturation, explanation of 34 

cycle, energy kxst in 34 

irreversible 34 

description of 32 

empirical equation for 48 

loop 33 

V8, heating 34 

separated from the eddy current loss 53 

measurement of 40 

* mechanical analogue to 36 

Induced e.m.f ., formulae for 57 

in a D.C. machine 75 

in an alternator and in an induction machine 65 

in a transformer 62 

Inductance and e.m.f ., relation between 185 



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INDEX 275 

PAGB 

Inductance as electromagnetic inertia 180 

definition of, and formulae for 184 

henry and perm as units of 184 

mechanical analogue of 185 

of a concentric cable, single phase 191 

of circuits in the presence of iron 186 

of coils and loops 188 

of coils in a D.C. machine 236 

of synchronous machines, measurement of, in A.C. machines 

219, 231 

of transformers, constants for 215 

formulae for 211-214 

vs. the end coils 213 

vs. the number of coils 213 

vs. the shape of the coils 212 

of transmission lines, single phase 199 

three-wire symmetrical spacing 201 

of windings, formula for 219 

fractional pitch 225 

how measured 219 

. vs. commutation 232 

vs. leakage permeance 184, 219 

Induction, law of 6, 57 

machines, armature reaction 131 

e.m.f . induced in 65 

higher harmonics in the m.m.f. of a 136 

inductance vs. winding pitch 225 

leakage flux, description of the 221 

leakage permeance, values of the 225 

motor, ratio of transformation in a 134 

secondary current in a 132 

torquein 259-260 

of e.m.f ., methods of 55 

Inertia, electromagnetic 180 

Insulator, magnetic 17 

Intensity, magnetic, definition of 13 

Interference. See Armature reaction, and Reaction. 

Interpoles, definition of {See also Computating poles) 172 

Irregular paths, permeance of; how to map field in 116 

Iron, grades of and their use 22 

laminations, preparation of 50 

reason for 41 

loss, definition of 42 

effects of 43 

figure of loss 47 

magnetization curves 20-24 

properties of, general 20, 32 



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276 INDEX 

PAGE 

Iron, saturation curves 20-24 

silicon steel 49 

space factor in laminations 41 

used in permanent magnets 60 

virgin state of 32 

V8. inductance 186 

wire cores 41 

Joints in transformers 81 

Joulecen, definition of 264 

Kelvin's law 253 

Knee of saturation curve 25 



ations, grades of 44 

preparation of 50 

reason for 41 

Lateral compression in the magnetic field 244 

Law of induction 6, 57 

Law, Ohm's, for magnetic circuit 7 

Laws of circulation 7, 57 

Leakage coil, definition of 215 

factor, calculation of 108 

definition of 107 

V8. leakage flux and permeance 109 

Leakage flux 16 

about armature windings 218 

belt, description of 223 

in field poles, as affected by load 113 

as affected by saturation 112 

effect of 108 

in induction motors, description of 221 

in transformers .' 208 

nature of, in machinery 86 

zig-zag, description of . 223 

Leakage inductance. See Inductance. 

See also Leakage permeance. 

Leakage permeance between field poles 108 

in induction machines, values of 225 

in synchronous machines, values of 230 

of coils in a D.C. machine 236 

of slots, calculation of 226 

of windings, formula for 220 

of zig-zag or tooth tip leakage, calculation of 227 

Lehmann, Dr. Th., method of finding permeance of irregular field 18 

Lifting magnet 243 



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INDEX 277 

PAGE 

Lines of force 2 

Linkages, fact of 3 

as a measure of energy 180 

partial and complete, definition of 181 

Load characteristics of alternator, definition of 141 

Loaded D.C. machine, flux distribution in 169 

field m.m.f . in 170 

D.C. motor, speed of a 171 

induction machine, m.m.f. relations in 132 

synchronous machine, flux distribution in 139 

Loading, specific electric, definition of 163 

Longitudinal tension in magnetic field 242 

Loop, hysteresis 33 

Machinery, core loss in revolving 46 

leakage flux in, nature of 86 

Machines, revolving, how torque is produced in 255, 258 

torque in 259, 260 

types of sjrnchronous 63 

Magnets, A.C., average tractive effort in 254 

average tractive effort in D.C 252 

lifting 243 

molecular 34 

permanent, iron used in 50 

torque of rotary magnets 253 

tractive, force of a 248 

Magnetic circuit, definition of 3 

Ohm's law for the 7 

types of, in revolving machinery 85 

simple 1 

with iron, in series and parallel 29 

Magnetic field, description of 1 

density of energy in 240 

energy stored in, description of 177 

formiilae for 181 

formulae for the actual force in a 251 

formulae for average force in a 252 

lateral compression in 244 

longitudinal tension in 243 

of transmission line, description of 193 

shape of 196 

reason for mechanical forces in 242 

Magnetic insulation 17 

intensity, definition of 13 

relation to m.m.f 13 

potential 16 

state 1 



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278 INDEX 

PAGE 

Magnetization curves. {See also Saturation curves.) 20-24 

Magnetizing current V8, exciting current 80 

Magnetism, cause of ... 1,4 

M.m.f . or magneto motive force, definition of 5 

for air-gap 89 

for armature cores 105 

fOT a D.C. machine to compensate for distortion 168 

for field frames 106^ 

for field poles 107 

for teeth, saturated 101 

for teeth, tapered 100 

gliding 126 

in induction machines, higher harmonics of 136 

of field, in a loaded D. C. machine 170 

of windings, concentrated 121 

distributed single phase. 123 

distributed polyphase 128 

relations in an induction machine 132 

a sjnichronous machine 143 

wave, harmonics of 123, 136 

Maxwell, definition of . . 6 

Measurement of core loss 44 

of hysteresis 40 

of inductance 219 

Mechanical analogue of hysteresis 36 

of inductance 185 

force. See Force, mechanical. 

Methods of inducing e.m.f 55 

Molecular magnets ' 34 

Multiplex windings, commutation of 238 

Mutual induction, in transmission lines 205 

in the windings of machines 219 

See also Transformer action 55 

Neutral zone, in D.C. machine * 163 

Ohm's law for the magnetic circuit . . . . : 7 

Open circuit characteristic. See Saturation curves. 

current. See Exciting current. 
Overload capacity of a synchronous motor 148 

Parallel combination of permeances 16 

-series circuits containing iron 29 

Partial linkages, definition of 181 

Perm, definition of 9 

Permeability curves 24 

definition of 11 



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INDEX 279 

PAGX 

Permeability, equation 24 

of air, discussed 266 

of non-magnetic materials . 11 

relative 24 

V8, reluctivity 11 

V8. saturation 24 

Permeance, combinations of, in parallel and series 16 

definition of ; 9 

equivalent, definition of 184 

leakage, in D.C. machines 236 

in induction machines, values of 225 

in synchronous machines, values of 230 

measurement of 219, 231 

of slot, formula for 226 

of windings, formula for 220 

zigzag, formula for 228 

of air-gap, accurate method 92-97 

simplified 89 

of irregular paths 116 

of pole fringe 94 

of tooth fringe 93 

units of 9 

vs. dimensions 11 

Phase characteristics of a sjmchronous motor 142 

relation of m.m.fs. in an induction machine 132 

in a synchronous machine 145 

Pinch phenomenon 246 

Pole face windings, Ryan 174 

fringe permeance 94 

Poles, fictitious, assumed in synchronous machines 151 

non-salient in a synchronous machine 121, 143 

salient, definition of 142 

Potential, definition of 16 

Potier diagram 146 

Pulsating- and gliding m.m.fs 126 

Ratio of transformation 134 

of voltages in a rotary converter 78 

Rayleigh, Lord, method for finding permeance of a field 116 

Reactance, equivalent, of an unsymmetrical transmission line 206 

in s3mchronous machines, nature of 140 

voltage, average, as criterion of commutation 235 

Reaction. See also Armature reaction. 

demagnetizing, in a D.C. machine > 164 

direct and transverse, nature of 150 

calculation of the coefficient of 158 

definition of the coefficient of 153 



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280 INDEX 

FAQS 

Reaction, distorting. See Transverse reaction. 

transverse, calculation of the coefficient of 160 

definition of the coefficient of 153 

in a direct current machine 165 

in a synchronous machine 150 

Rectangular m.m.f. wave, harmonics of 123 

Regulation of alternators, calculation of 146-8, 155--6 

definition of 156 

Rel, definition of 8 

Relative permeability 24 

Reluctance, definition of 7 

in series and in parallel 16 

of irregular paths 116 

of various magnetic circuits. See Permeance. 

unit of 8 

V8. dimensions 11 

Reluctivity, definition of 11 

of air ; 11 

vs. permeability 11 

Remanent magnetism. See Residual. 

Repulsion between bus-bars 248 

between transformer coils 245 

Residual magnetism 32 

Resistance, equivalent, of.unsymmetrical transmission line 207 

Revolving field 126 

machinery, core loss in 46 

types of magnetic circuit in 85 

See also particular kind of machinery. 

Right-hand screw rule 2 

Rotary converter, armature reaction in a 175 

voltage ratio in a 78 

voltage regulation in a 78 

Ryan, pole face winding for D.C. machines 174 

Salient poles, definition of 142 

in sjnichronous machines, Blondel diagram for 154 

Saturation and hysteresis, an explanation of 34 

and permeability 24 

curve, knee of 25 

curves of iron 20-24 

of machinery. See also Transformers, exciting 

current 85-87 

effect of, upon pole leakage 112 

per cent of 26 

Secondary current, calculation of, in an induction motor 132 

winding, equivalent, reduced to the primary 133 

Semi-net length 220 



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INDEX 281 

PAQB 

Semi-symmetrical spacing 203 

Series-parallel circuits with iron 29 

•Series-reluctances 16 

Sheet metal. See Laminations. 

Short chord or short pitch windings. See Fractional pitch windings. 

Silicon steel 49 

Simple magnetic circuit 1 

Skin effect 188 

Slot factor, definition of 68 

formulae for 69 

leakage permeance, calculation of 226 

Slotted armature, how torque is produced in 258 

Solenoidal, term applied to magnetic field 6 

Space factor in iron 41 

Sparking. See Commutation. 

Speed of a variable-speed D.C. motor under load 171 

Squirel-cage winding 133 

Steinmetz's law for hysteresis 48 

Steel. See Iron. 

Stray flux or leakage 16 

Superposition, principle of 194 

S3nichronous condensers 157 

Sjmchronous machines, armature reaction in 140 

e.m.f . induced in 65 

fictitious poles assumed in 151 

loaded, flux distribution in 139 

measurement of leakage inductance 219, 231 

overload capacity of motor 148 

phase relation of m.m.fs. in a 143-145 

reactive drop in a Ginodts of) 229 

torque in 259, 260 

types of 63 

values of leakage permeance of the windings of. . 230 
vector diagrams of, with non-salient poles . . 145, 147 

with salient poles 154 

Sjnichronous motor, definition of phase characteristices of 142 

overload capacity of J48 

Teeth, apparent flux density in 101 

permeance of tooth fringe 93 

saturated, m.m.f . for 101 

tapered, m.m.f. for 100 

Tension. See also Force, mechanical 

longitudinal, in magnetic field 242 

Torque. See also Force, mechanical. 

from equivalent electric circuit 260 

in revolving machines, how produced 258 



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282 INDEX 

PAOB 

Torque in slotted armatures, how produced 258 

of rotary magnets 263 

of revolving machinery 268- 260 

Tractive effort, actual .261 

average, in a D.C. magnet 252 

in an A.C. magnet, average 254 

magnet, force of 248 

Transformer action 66 

Transformers, core loss current in a 82 

exciting current in a 80, 83 

induced e.m.f. in a 62 

inductance of 211-214 

as afifected by end coils 213 

as afifected by number of coils , . 213 

as afifected by the shape of the coils 212 

values of the constants 215 

joints in the magnetic circuit of a 81 

leakage flux in a 208 

magnetizing current in a 80 

repulsion between the coils of a 245 

types of 60 

Transmission line, description of the field of a 193 

shape of the field calculated 196 

single phase, inductance of 199 

three-wire, symmetrical spacing, inductance of 201 

unsymmetrical spacing, e.m.f. induced in 205 

unsymmetrical spacing, equivalent reactance of 206 

unsymmetrical spacing, equivalent resistance of 207 

Transverse and direct reaction, nature of 150 

reaction, calculation of the coefficient of 160 

definition of the coefficient of 153 

in a D.C. machine 165 

in a synchronous machine. 150 

Turning moment in machinery .258-260 

Unit of flux 6 

of inductance 184 

of permeance 9 

of reluctance 8 

Units, table of the dimensions of 263 

V curve or phase characteristics 142 

Variable-speed D.C. motor under load 171 

Vector relations in a synchronous machine 145, 147, 164, 155 

with non-salient poles 145 

with salient poles 154 

Virgin state of iron 32 



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INDEX 283 

PAGB 

Virtual displacements, principle of 249 

Voltage. See E.m.f. 

Wave form of e.m.f., effect of type of windings upon 73 

how to get sine wave 72 

Wave-wound armatures, commutation of 237 

Weber, definition of 6 

Wire, iron, for cores 41 

Winding-pitch factor, values of 71 

Windings, compensating, for D.C. machines, 174 

concentrated, m.m.f . of 121 

for D.C. machines, effect of type on voltage 76 

distributed, definition of 121 

in A.C. machines, advantages and disadvantages 

of 66 

formula for the inductance of 219 

fractional pitch, advantages and disadvantages of 67 

conmiutation of 238 

effect of, on induced e.m.f 67, 76 

on inductance 225, 231 

inductance and reactance, how measured 219 

leakage permeance of, formula for 220 

multiplex, commutation of 238 

pole-face for D.C. machines 174 

polyphase, m.m.f. of 128 

wave or two-circuit, conmiutation of 237 

Ymeh, definition of 10 

Zig-zag leakage, description of 223 

permeance, calculation of 227 



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